Chapter 4 Determinants (Additional Questions)

To find the value of a $2 \times 2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, we use the formula $ad - bc$.

In the given determinant $\begin{vmatrix} 2 & -3 \\ 4 & 5 \end{vmatrix}$, we have $a=2$, $b=-3$, $c=4$, and $d=5$.

Value of the determinant $= (2)(5) - (-3)(4)$

$= 10 - (-12)$

$= 10 + 12$

$= 22$

The value of the determinant is $22$.

Thus, the correct option is (C).

Given that $A$ is a square matrix of order $3 \times 3$.

Also, the determinant of matrix $A$ is given as $|A| = 5$.

We need to find the value of $|A'|$, where $A'$ is the transpose of matrix $A$.

A fundamental property of determinants is that the determinant of a matrix is equal to the determinant of its transpose. This property holds true for any square matrix.

Mathematically, this property is stated as:

$|A'| = |A|$

Using the given value of $|A|$, we can find $|A'|$.

$|A'| = 5$

Thus, the value of $|A'|$ is $5$.

The correct option is (A).

The vertices of the triangle are given as $(1, 0)$, $(4, 0)$, and $(4, 3)$.

The area of the triangle is given by the absolute value of $\frac{1}{2}$ times the determinant of the matrix formed by the coordinates of the vertices and a column of ones:

Area $= \left| \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \right|$

Substituting the given vertices, the determinant is $\begin{vmatrix} 1 & 0 & 1 \\ 4 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}$, as provided in the question.

Let's calculate the value of the determinant $\Delta = \begin{vmatrix} 1 & 0 & 1 \\ 4 & 0 & 1 \\ 4 & 3 & 1 \end{vmatrix}$.

We can expand the determinant along the first row:

$\Delta = 1 \times \text{cofactor}(a_{11}) + 0 \times \text{cofactor}(a_{12}) + 1 \times \text{cofactor}(a_{13})$

$\Delta = 1 \times (-1)^{1+1} \begin{vmatrix} 0 & 1 \\ 3 & 1 \end{vmatrix} + 0 \times (-1)^{1+2} \begin{vmatrix} 4 & 1 \\ 4 & 1 \end{vmatrix} + 1 \times (-1)^{1+3} \begin{vmatrix} 4 & 0 \\ 4 & 3 \end{vmatrix}$

$\Delta = 1 \times (0 \times 1 - 1 \times 3) + 0 + 1 \times (4 \times 3 - 0 \times 4)$

$\Delta = 1 \times (0 - 3) + 1 \times (12 - 0)$

$\Delta = 1 \times (-3) + 1 \times (12)$

$\Delta = -3 + 12$

$\Delta = 9$

The value of the determinant is $9$.

Now, we find the area of the triangle using the given formula:

Area $= \left| \frac{1}{2} \times \Delta \right|$

Area $= \left| \frac{1}{2} \times 9 \right|$

Area $= \left| \frac{9}{2} \right|$

Area $= 4.5$ square units.

The calculated area of the triangle based on the provided determinant is $4.5$ square units.

Please note that $4.5$ is not one of the given options (A) $6$, (B) $3$, (C) $12$, (D) $0$.

However, if the vertices were meant to form a right triangle with base 4 and height 3 (for example, vertices (0,0), (4,0), (4,3) or (0,0), (3,0), (3,4)), the area would be $\frac{1}{2} \times 4 \times 3 = 6$ or $\frac{1}{2} \times 3 \times 4 = 6$, which matches option (A).

But strictly following the calculation using the determinant provided for the vertices $(1, 0), (4, 0), (4, 3)$, the area is $4.5$.

This question refers to a fundamental property of determinants.

One of the key properties of determinants states that if any two rows or any two columns of a determinant are identical (meaning they have the same elements in the same order), then the value of the determinant is zero.

Let's illustrate with a $2 \times 2$ example:

Consider the determinant $\begin{vmatrix} a & b \\ a & b \end{vmatrix}$ where the first row is identical to the second row.

The value of this determinant is calculated as $(a)(b) - (b)(a) = ab - ab = 0$.

Similarly, for a $3 \times 3$ determinant with identical rows or columns, the value will also be $0$.

Therefore, if any two rows or columns of a determinant are identical, the value of the determinant is $\mathbf{0}$.

The correct option is (C).

Given matrix $A = \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}$.

For a $2 \times 2$ matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the adjoint of $M$ is given by interchanging the elements on the main diagonal and changing the signs of the off-diagonal elements.

The formula for $adj(M)$ is $\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

In the given matrix $A$, we have $a=2$, $b=1$, $c=3$, and $d=4$.

Applying the formula, the adjoint of $A$ is:

$adj(A) = \begin{bmatrix} 4 & -(1) \\ -(3) & 2 \end{bmatrix}$

$adj(A) = \begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}$

The adjoint of matrix $A$ is $\begin{bmatrix} 4 & -1 \\ -3 & 2 \end{bmatrix}$.

This matches option (A).

Given that $A$ is an invertible matrix of order $2 \times 2$.

For an invertible matrix $A$, its inverse $A^{-1}$ exists, and the product of the matrix and its inverse is the identity matrix $I$ of the same order.

$A A^{-1} = I$

Taking the determinant of both sides of the equation:

$|A A^{-1}| = |I|$

Using the property that the determinant of a product of matrices is the product of their determinants, i.e., $|AB| = |A||B|$, we have:

$|A| |A^{-1}| = |I|$

The determinant of an identity matrix of any order is $1$. So, $|I| = 1$.

$|A| |A^{-1}| = 1$

Since $A$ is invertible, its determinant $|A|$ must be non-zero ($|A| \neq 0$). Therefore, we can divide both sides by $|A|$:

$|A^{-1}| = \frac{1}{|A|}$

Thus, the determinant of the inverse of an invertible matrix is the reciprocal of the determinant of the original matrix.

The value of $|A^{-1}|$ is equal to $\frac{1}{|A|}$.

The correct option is (B).

We are asked to find the value of the determinant $\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$.

We can calculate the determinant by expanding along the first row (or any other row or column).

Determinant $= 1 \times \text{cofactor}(a_{11}) - 2 \times \text{cofactor}(a_{12}) + 3 \times \text{cofactor}(a_{13})$

Determinant $= 1 \times (-1)^{1+1} \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} + 2 \times (-1)^{1+2} \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \times (-1)^{1+3} \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}$

Determinant $= 1 \times (5 \times 9 - 6 \times 8) - 2 \times (4 \times 9 - 6 \times 7) + 3 \times (4 \times 8 - 5 \times 7)$

Determinant $= 1 \times (45 - 48) - 2 \times (36 - 42) + 3 \times (32 - 35)$

Determinant $= 1 \times (-3) - 2 \times (-6) + 3 \times (-3)$

Determinant $= -3 + 12 - 9$

Determinant $= 9 - 9$

Determinant $= 0$

Alternatively, we can observe the relationship between the rows.

Let $R_1, R_2, R_3$ be the first, second, and third rows respectively.

$R_2 - R_1 = (4-1, 5-2, 6-3) = (3, 3, 3)$

$R_3 - R_2 = (7-4, 8-5, 9-6) = (3, 3, 3)$

Since the difference between consecutive rows is the same, the rows are in arithmetic progression. This implies that the rows are linearly dependent (specifically, $R_1 - 2R_2 + R_3 = 0$).

A property of determinants states that if the rows or columns of a matrix are linearly dependent, the value of the determinant is $\mathbf{0}$.

Thus, the value of the determinant is $0$.

The correct option is (B).

This question is based on a fundamental property of square matrices and their adjoints.

For any square matrix $A$ of order $n$, the product of the matrix $A$ and its adjoint $adj(A)$ is equal to the product of the determinant of $A$ and the identity matrix $I$ of the same order $n$.

This property can be written as:

$A (adj A) = (adj A) A = |A| I$

where $|A|$ is the determinant of matrix $A$, and $I$ is the identity matrix of the same order as $A$.

Therefore, for any square matrix $A$, $A(adj A)$ is equal to $|A| I$.

The correct option is (C).

Given that the points $(x, -2)$, $(5, 2)$, and $(8, 8)$ are collinear.

When three points are collinear, the area of the triangle formed by these points is $0$.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the absolute value of $\frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$.

Since the area is $0$, the determinant $\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix}$ must be $0$.

Substituting the given points $(x, -2)$, $(5, 2)$, and $(8, 8)$ into the determinant, we get:

$\begin{vmatrix} x & -2 & 1 \\ 5 & 2 & 1 \\ 8 & 8 & 1 \end{vmatrix} = 0$

Now, we expand the determinant, for example, along the first row:

$x \begin{vmatrix} 2 & 1 \\ 8 & 1 \end{vmatrix} - (-2) \begin{vmatrix} 5 & 1 \\ 8 & 1 \end{vmatrix} + 1 \begin{vmatrix} 5 & 2 \\ 8 & 8 \end{vmatrix} = 0$

$x ((2)(1) - (1)(8)) + 2 ((5)(1) - (1)(8)) + 1 ((5)(8) - (2)(8)) = 0$

$x (2 - 8) + 2 (5 - 8) + 1 (40 - 16) = 0$

$x (-6) + 2 (-3) + 1 (24) = 0$

$-6x - 6 + 24 = 0$

$-6x + 18 = 0$

$-6x = -18$

$x = \frac{-18}{-6}$

$x = 3$

The value of $x$ that makes the points collinear is $3$.

The correct option is (C).

Given matrix $A = \begin{bmatrix} 3 & 1 \\ 2 & 5 \end{bmatrix}$.

We need to find the minor of the element $a_{21}$.

The element $a_{21}$ is the element in the 2nd row and 1st column of the matrix $A$. In this case, $a_{21} = 2$.

The minor of an element $a_{ij}$, denoted by $M_{ij}$, is the determinant of the submatrix obtained by deleting the $i$-th row and $j$-th column of the original matrix.

To find the minor of $a_{21}$ ($M_{21}$), we delete the 2nd row and the 1st column of matrix $A$:

Original matrix: $\begin{bmatrix} 3 & 1 \\ \cancel{2} & \cancel{5} \end{bmatrix}$

After deleting the 2nd row and 1st column, the remaining submatrix is $[1]$.

The minor $M_{21}$ is the determinant of this submatrix:

$M_{21} = \det([1]) = 1$

The minor of the element $a_{21}$ is $1$.

The correct option is (B).

Let's examine each statement:

(A) It is defined for any matrix.

This statement is false. The determinant is only defined for square matrices, i.e., matrices with the same number of rows and columns.

(B) It is a scalar value.

This statement is true. The determinant is a single numerical value calculated from the elements of a square matrix.

(C) It changes sign when two rows are interchanged.

This statement is true. This is a property of determinants. If two rows (or two columns) of a determinant are interchanged, the value of the determinant changes its sign.

For example, $\begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc$, but $\begin{vmatrix} c & d \\ a & b \end{vmatrix} = cb - da = -(ad - bc)$.

(D) The determinant of an identity matrix is 0.

This statement is false. The identity matrix, denoted by $I$ or $I_n$ for an $n \times n$ matrix, has ones on the main diagonal and zeros elsewhere. The determinant of an identity matrix of any order is always $1$.

For example, $|I_2| = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = (1)(1) - (0)(0) = 1$.

$|I_3| = \begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} = 1 \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} - 0 + 0 = 1(1-0) = 1$.

Based on the analysis, the true statements are (B) and (C).

Let's analyze the Assertion (A) and the Reason (R).

Assertion (A): If $|A| = 0$, then $A^{-1}$ does not exist.

This statement is a fundamental property of matrices. A square matrix $A$ is invertible (or non-singular) if and only if its determinant, $|A|$, is non-zero. If the determinant $|A|$ is zero, the matrix is called singular, and a singular matrix does not have an inverse. So, Assertion (A) is true.

Reason (R): $A^{-1} = \frac{1}{|A|} adj(A)$.

This is the standard formula for calculating the inverse of a square matrix $A$, where $adj(A)$ is the adjoint of $A$ and $|A|$ is the determinant of $A$. This formula is valid provided that $|A| \neq 0$. So, Reason (R) is a true statement (it is the formula for the inverse when the inverse exists).

Now, let's check if Reason (R) is the correct explanation for Assertion (A).

The formula $A^{-1} = \frac{1}{|A|} adj(A)$ shows that the calculation of the inverse requires division by the determinant $|A|$.

If $|A| = 0$, the formula becomes $A^{-1} = \frac{1}{0} adj(A)$.

Division by zero is undefined. Therefore, if $|A|=0$, the expression $\frac{1}{|A|} adj(A)$ is undefined, which means $A^{-1}$ does not exist.

The Reason (R) provides the formula for the inverse, which explicitly contains the determinant in the denominator, thereby explaining why the inverse does not exist when the determinant is zero.

Thus, Reason (R) is the correct explanation for Assertion (A).

Both the Assertion and the Reason are true, and the Reason correctly explains the Assertion.

The correct option is (A).

Given that $A$ is a square matrix of order $n$, and its determinant $|A| \neq 0$.

We know the property relating a square matrix, its adjoint, and its determinant:

$A (adj A) = (adj A) A = |A| I_n$

where $I_n$ is the identity matrix of order $n$.

Let's consider the equation $A (adj A) = |A| I_n$.

Taking the determinant of both sides, we get:

$|A (adj A)| = ||A| I_n|$

Using the property that the determinant of a product of matrices is the product of their determinants, $|AB| = |A| |B|$, the left side becomes:

$|A| |adj A|$

The right side is the determinant of a scalar multiple of the identity matrix. For a scalar $k$ and an $n \times n$ identity matrix $I_n$, $|k I_n| = k^n |I_n| = k^n \times 1 = k^n$.

Here, the scalar is $|A|$. So, $||A| I_n| = (|A|)^n$.

Equating the left and right sides, we have:

$|A| |adj A| = |A|^n$

Since it is given that $|A| \neq 0$, we can divide both sides by $|A|$:

$|adj A| = \frac{|A|^n}{|A|}$

$|adj A| = |A|^{n-1}$

Thus, the value of $|adj(A)|$ is $|A|^{n-1}$.

The correct option is (B).

Given that $A$ is a square matrix of order $n = 3 \times 3$.

The determinant of matrix $A$ is given as $|A| = 4$.

We need to find the value of $|3A|$, where $3$ is a scalar multiplying the matrix $A$.

There is a property of determinants that states: If $A$ is a square matrix of order $n$ and $k$ is any scalar, then $|kA| = k^n |A|$.

In this problem, the order of the matrix is $n=3$, and the scalar is $k=3$. The determinant of $A$ is $|A|=4$.

Applying the property, we have:

$|3A| = 3^n |A|$

$|3A| = 3^3 |A|$

Now, we calculate $3^3$:

$3^3 = 3 \times 3 \times 3 = 9 \times 3 = 27$

Substitute the value of $3^3$ and the given value of $|A|$ into the equation:

$|3A| = 27 \times 4$

Performing the multiplication:

$27 \times 4 = 108$

$|3A| = 108$

Thus, the value of $|3A|$ is $108$.

The correct option is (C).

Given that $A$ and $B$ are square matrices of the same order.

We need to find the value of the determinant of the product of matrices $A$ and $B$, which is $|AB|$.

There is a fundamental property of determinants that states that the determinant of the product of two square matrices of the same order is equal to the product of their individual determinants.

This property can be expressed mathematically as:

$|AB| = |A| |B|$

where $|A|$ is the determinant of matrix $A$ and $|B|$ is the determinant of matrix $B$.

Therefore, if $A$ and $B$ are square matrices of the same order, then $|AB|$ is equal to $|A| |B|$.

The correct option is (C).

The question asks for the condition under which a system of linear equations represented as $AX = B$ has a unique solution, where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

Given: A system of linear equations $AX = B$.

To Find: The condition for a unique solution.

Options:

(A) $|A| = 0$

(B) $|A| \neq 0$

(C) $adj(A)B = O$

(D) $adj(A)B \neq O$

Solution:

For a system of linear equations $AX = B$, where A is a square matrix, the nature of the solution (unique, infinite, or no solution) depends on the determinant of the coefficient matrix A, denoted as $|A|$.

Case 1: Unique Solution

A unique solution exists if and only if the determinant of the coefficient matrix A is non-zero.

When $|A| \neq 0$, the matrix A is invertible, and the solution can be found using the formula $X = A^{-1}B$.

Case 2: Non-Unique or No Solution

If $|A| = 0$, the system either has infinitely many solutions or no solution. In this case, further investigation using the augmented matrix or properties of the adjoint matrix is required.

Let's consider the options in relation to this principle:

(A) $|A| = 0$: This condition implies either no solution or infinitely many solutions, not a unique solution.

(B) $|A| \neq 0$: This is the standard condition for a unique solution in a system of linear equations.

(C) $adj(A)B = O$: This condition is related to cases where $|A|=0$. If $|A|=0$ and $adj(A)B = O$, the system may have infinitely many solutions. If $|A|=0$ and $adj(A)B \neq O$, the system has no solution.

(D) $adj(A)B \neq O$: This condition, when $|A|=0$, indicates that the system has no solution.

Therefore, the condition for a unique solution is that the determinant of the coefficient matrix A must be non-zero.

Final Answer:

The condition for a system of linear equations $AX = B$ to have a unique solution is $|A| \neq 0$.

The correct option is (B).

The question provides a 2x2 matrix A and states that its determinant $|A|$ is equal to 0. We need to find the value of $x$ in the matrix A.

Given:

Matrix $A = \begin{bmatrix} 1 & 2 \\ x & 4 \end{bmatrix}$

Condition: $|A| = 0$

To Find: The value of $x$.

Options:

(A) $2$

(B) $4$

(C) $-2$

(D) $0$

Solution:

The determinant of a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is calculated as $ad - bc$.

For the given matrix $A = \begin{bmatrix} 1 & 2 \\ x & 4 \end{bmatrix}$, the determinant is:

We are given that $|A| = 0$. So, we can set the expression for the determinant equal to 0:

Now, we solve this equation for $x$:

$4 - 2x = 0$

Add $2x$ to both sides:

$4 = 2x$

Divide both sides by 2:

$\frac{4}{2} = x$

$2 = x$

So, the value of $x$ is 2.

Final Answer:

The value of $x$ is 2.

The correct option is (A).

The question asks to find the cofactor of the element $a_{12}$ in the given 3x3 matrix.

Given Matrix:

$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$

To Find: The cofactor of the element $a_{12}$.

Options:

(A) $-6$

(B) $6$

(C) $-12$

(D) $12$

Solution:

The element $a_{12}$ is the element in the 1st row and 2nd column of the matrix A.

In the given matrix, $a_{12} = 2$.

The cofactor of an element $a_{ij}$ in a matrix is given by $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

The minor $M_{ij}$ is the determinant of the submatrix obtained by deleting the $i$-th row and the $j$-th column of the original matrix.

For the element $a_{12}$, we need to find $C_{12}$.

Here, $i=1$ and $j=2$.

First, let's find the minor $M_{12}$ by deleting the 1st row and 2nd column of matrix A:

$A = \begin{bmatrix} \cancel{1} & \cancel{2} & \cancel{3} \\ 4 & \cancel{5} & 6 \\ 7 & \cancel{8} & 9 \end{bmatrix}$

The submatrix is $\begin{bmatrix} 4 & 6 \\ 7 & 9 \end{bmatrix}$.

The minor $M_{12}$ is the determinant of this submatrix:

Calculating the determinant:

$M_{12} = (4 \times 9) - (6 \times 7)$

$M_{12} = 36 - 42$

$M_{12} = -6$

Now, we can find the cofactor $C_{12}$ using the formula $C_{ij} = (-1)^{i+j}M_{ij}$:

$C_{12} = (-1)^{1+2}M_{12}$

$C_{12} = (-1)^{3}(-6)$

$C_{12} = (-1)(-6)$

$C_{12} = 6$

Thus, the cofactor of the element $a_{12}$ is 6.

Final Answer:

The cofactor of the element $a_{12}$ is 6.

The correct option is (B).

The question states that the area of a triangle with given vertices is 4 square units, and we need to find the possible values of $k$.

Given:

Vertices of the triangle:

• $(x_1, y_1) = (k, 0)$
• $(x_2, y_2) = (4, 0)$
• $(x_3, y_3) = (0, 2)$

Area of the triangle = 4 square units.

To Find: The possible value(s) of $k$.

Options:

(A) $0$

(B) $8$

(C) $0$ or $8$

(D) $4$ or $-4$

Solution:

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by the formula:

Substitute the given coordinates and the area into the formula:

$4 = \frac{1}{2} |k(0 - 2) + 4(2 - 0) + 0(0 - 0)|$

Simplify the expression inside the absolute value:

$4 = \frac{1}{2} |k(-2) + 4(2) + 0|$

$4 = \frac{1}{2} |-2k + 8|$

Multiply both sides by 2:

$8 = |-2k + 8|$

The absolute value means that the expression inside can be either positive or negative.

Case 1: $-2k + 8 = 8$

Subtract 8 from both sides:

$-2k = 8 - 8$

$-2k = 0$

Divide by -2:

$k = \frac{0}{-2}$

$k = 0$

Case 2: $-2k + 8 = -8$

Subtract 8 from both sides:

$-2k = -8 - 8$

$-2k = -16$

Divide by -2:

$k = \frac{-16}{-2}$

$k = 8$

So, the possible values of $k$ are 0 and 8.

Final Answer:

The possible values of $k$ are 0 and 8.

The correct option is (C).

The question asks to find the inverse of a matrix $A$, given its adjoint matrix ($adj(A)$) and its determinant ($|A|$).

Given:

Adjoint matrix, $adj(A) = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}$

Determinant of matrix A, $|A| = 5$

To Find: The inverse of matrix A, $A^{-1}$.

Options:

(A) $\begin{bmatrix} 10 & 15 \\ 20 & 5 \end{bmatrix}$

(B) $\begin{bmatrix} 2/5 & 3/5 \\ 4/5 & 1/5 \end{bmatrix}$

(C) $\begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}$

(D) $\begin{bmatrix} 1/5 & 0 \\ 0 & 1/5 \end{bmatrix}$

Solution:

The formula for the inverse of a matrix $A$ is given by:

We are given $|A| = 5$ and $adj(A) = \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}$.

Substitute these values into the formula for $A^{-1}$:

$A^{-1} = \frac{1}{5} \begin{bmatrix} 2 & 3 \\ 4 & 1 \end{bmatrix}$

To multiply a scalar with a matrix, we multiply each element of the matrix by the scalar:

$A^{-1} = \begin{bmatrix} \frac{1}{5} \times 2 & \frac{1}{5} \times 3 \\ \frac{1}{5} \times 4 & \frac{1}{5} \times 1 \end{bmatrix}$

$A^{-1} = \begin{bmatrix} \frac{2}{5} & \frac{3}{5} \\ \frac{4}{5} & \frac{1}{5} \end{bmatrix}$

Now, let's compare this result with the given options.

Option (A) is $\begin{bmatrix} 10 & 15 \\ 20 & 5 \end{bmatrix}$, which is $5 \times adj(A)$, not $\frac{1}{5} \times adj(A)$.

Option (B) is $\begin{bmatrix} 2/5 & 3/5 \\ 4/5 & 1/5 \end{bmatrix}$, which matches our calculated $A^{-1}$.

Option (C) is $adj(A)$ itself.

Option (D) is $\begin{bmatrix} 1/5 & 0 \\ 0 & 1/5 \end{bmatrix}$, which is $\frac{1}{5}I$, where I is the identity matrix.

Final Answer:

The inverse of the matrix A is $\begin{bmatrix} 2/5 & 3/5 \\ 4/5 & 1/5 \end{bmatrix}$.

The correct option is (B).

The question asks to identify the correct properties of a determinant from the given options.

Options:

(A) If a scalar is multiplied to a determinant, only one row or one column is multiplied by the scalar.

(B) The value of the determinant remains unchanged if a row operation $R_i \to R_i + k R_j$ is performed.

(C) The determinant of a diagonal matrix is the product of its diagonal elements.

(D) $|A+B| = |A| + |B|$ for square matrices $A$ and $B$ of the same order.

Analysis of each option:

(A) If a scalar is multiplied to a determinant, only one row or one column is multiplied by the scalar.

This statement is correct. If a scalar $k$ is multiplied by a determinant, it means multiplying each element of *one* row or *one* column by $k$. For example, if we have a matrix $A$, then $det(kA)$ is not $k \cdot det(A)$ (unless A is 1x1). Instead, if $A$ is an $n \times n$ matrix, $det(kA) = k^n \cdot det(A)$. The property mentioned in option A is about multiplying the determinant value by a scalar, which is equivalent to multiplying one row or column by that scalar.

Let's verify with a 2x2 example: $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $det(A) = ad-bc$.

If we multiply the first row by $k$: $\begin{bmatrix} ka & kb \\ c & d \end{bmatrix}$. The determinant is $(ka)d - (kb)c = k(ad-bc) = k \cdot det(A)$.

(B) The value of the determinant remains unchanged if a row operation $R_i \to R_i + k R_j$ is performed.

This statement is correct. Adding a multiple of one row to another row does not change the value of the determinant.

(C) The determinant of a diagonal matrix is the product of its diagonal elements.

This statement is correct. For a diagonal matrix (where all off-diagonal elements are zero), the determinant is indeed the product of the elements on the main diagonal. This also applies to triangular matrices (upper or lower).

Example: $A = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$. $det(A) = ad - 0 \times 0 = ad$.

(D) $|A+B| = |A| + |B|$ for square matrices $A$ and $B$ of the same order.

This statement is incorrect. The determinant of the sum of two matrices is generally not equal to the sum of their determinants. This property only holds in specific cases, not as a general rule.

Example: Let $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$.

$|A| = 1 \times 0 - 0 \times 0 = 0$.

$|B| = 0 \times 1 - 0 \times 0 = 0$.

$|A| + |B| = 0 + 0 = 0$.

However, $A+B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ (Identity matrix).

$|A+B| = |I| = 1 \times 1 - 0 \times 0 = 1$.

Here, $|A+B| \neq |A| + |B|$.

Conclusion:

The correct properties are (A), (B), and (C).

Final Answer:

The correct properties for a determinant are:

(A) If a scalar is multiplied to a determinant, only one row or one column is multiplied by the scalar.

(B) The value of the determinant remains unchanged if a row operation $R_i \to R_i + k R_j$ is performed.

(C) The determinant of a diagonal matrix is the product of its diagonal elements.

The question asks to determine the nature of the solution for a given system of linear equations.

Given System of Linear Equations:

1. $x + 2y = 5$
2. $2x + 4y = 10$

To Find: The number of solutions for the system.

Options:

(A) A unique solution

(B) No solution

(C) Infinitely many solutions

(D) Exactly two solutions

Solution:

We can analyze the system of equations using their coefficients or by attempting to solve them.

Let's represent the system in matrix form $AX = B$, where $A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $B = \begin{bmatrix} 5 \\ 10 \end{bmatrix}$.

To determine the nature of the solution, we can look at the determinant of the coefficient matrix A:

$|A| = 4 - 4 = 0$

Since the determinant of the coefficient matrix is 0, the system will either have no solution or infinitely many solutions. To distinguish between these two cases, we can check the relationship between the equations.

Let's examine the second equation:

$2x + 4y = 10$

If we divide the entire second equation by 2, we get:

$\frac{2x}{2} + \frac{4y}{2} = \frac{10}{2}$

$x + 2y = 5$

This resulting equation is identical to the first equation ($x + 2y = 5$). This means that the two equations are dependent; they represent the same line.

When the equations in a system of linear equations are dependent, there are infinitely many points that satisfy both equations (as they are the same equation). Therefore, the system has infinitely many solutions.

Alternatively, using the condition for consistency:

For a system of equations $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, there are infinitely many solutions.

In our case:

• $a_1 = 1, b_1 = 2, c_1 = 5$
• $a_2 = 2, b_2 = 4, c_2 = 10$

Check the ratios:

$\frac{a_1}{a_2} = \frac{1}{2}$

$\frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2}$

$\frac{c_1}{c_2} = \frac{5}{10} = \frac{1}{2}$

Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the system has infinitely many solutions.

Final Answer:

The system of linear equations has infinitely many solutions.

The correct option is (C).

The question asks to match the concepts in Column I with their corresponding definitions or properties in Column II.

Column I: Concepts

1. (i) Minor of an element $a_{ij}$
2. (ii) Cofactor of an element $a_{ij}$
3. (iii) Singular matrix
4. (iv) Non-singular matrix

Column II: Definitions/Properties

1. (a) Determinant is non-zero
2. (b) Determinant of the submatrix obtained by deleting row $i$ and column $j$
3. (c) $(-1)^{i+j} \times (\text{Minor of } a_{ij})$
4. (d) Determinant is zero

Matching:

(i) Minor of an element $a_{ij}$:

The minor of an element $a_{ij}$ is defined as the determinant of the submatrix formed by deleting the $i$-th row and the $j$-th column of the original matrix.

This matches definition (b) in Column II.

Match: (i) $\rightarrow$ (b)

(ii) Cofactor of an element $a_{ij}$:

The cofactor of an element $a_{ij}$ is given by the formula $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor of $a_{ij}$.

This matches definition (c) in Column II.

Match: (ii) $\rightarrow$ (c)

(iii) Singular matrix:

A square matrix is called a singular matrix if its determinant is equal to zero.

This matches definition (d) in Column II.

Match: (iii) $\rightarrow$ (d)

(iv) Non-singular matrix:

A square matrix is called a non-singular matrix if its determinant is not equal to zero.

This matches definition (a) in Column II.

Match: (iv) $\rightarrow$ (a)

Final Answer:

The correct matches are:

• (i) $\rightarrow$ (b)
• (ii) $\rightarrow$ (c)
• (iii) $\rightarrow$ (d)
• (iv) $\rightarrow$ (a)

The question asks to find the determinant of a given 3x3 matrix $A$.

Given Matrix:

$A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{bmatrix}$

To Find: The determinant of matrix $A$, $|A|$.

Options:

(A) $11$

(B) $24$

(C) $0$

(D) $1$

Solution:

The given matrix $A$ is an upper triangular matrix because all the elements below the main diagonal are zero.

For any triangular matrix (either upper triangular or lower triangular), the determinant is equal to the product of its diagonal elements.

The diagonal elements of matrix $A$ are $a_{11} = 1$, $a_{22} = 4$, and $a_{33} = 6$.

Therefore, the determinant of $A$ is:

$|A| = 1 \times 4 \times 6$

$|A| = 24$

Alternatively, we can calculate the determinant using cofactor expansion along the first column:

$|A| = 1 \times C_{11} + 0 \times C_{21} + 0 \times C_{31}$

$|A| = 1 \times C_{11}$

The minor $M_{11}$ is the determinant of the submatrix obtained by removing the first row and first column:

$M_{11} = \begin{vmatrix} 4 & 5 \\ 0 & 6 \end{vmatrix} = (4 \times 6) - (5 \times 0) = 24 - 0 = 24$.

The cofactor $C_{11} = (-1)^{1+1} M_{11} = (-1)^2 \times 24 = 1 \times 24 = 24$.

So, $|A| = 1 \times 24 = 24$.

Final Answer:

The determinant of the matrix $A$ is 24.

The correct option is (B).

The determinant of the coefficient matrix $A = \begin{bmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 3 & 2 & -4 \end{bmatrix}$ needs to be calculated.

Using cofactor expansion along the first row:

$|A| = 1 \begin{vmatrix} -1 & 1 \\ 2 & -4 \end{vmatrix} - 1 \begin{vmatrix} 1 & 1 \\ 3 & -4 \end{vmatrix} + 1 \begin{vmatrix} 1 & -1 \\ 3 & 2 \end{vmatrix}$

$|A| = 1 [(-1)(-4) - (1)(2)] - 1 [(1)(-4) - (1)(3)] + 1 [(1)(2) - (-1)(3)]$

$|A| = 1 [4 - 2] - 1 [-4 - 3] + 1 [2 + 3]$

$|A| = 1 [2] - 1 [-7] + 1 [5]$

$|A| = 2 + 7 + 5$

$|A| = 14$

The calculated determinant is 14. However, 14 is not among the given options (A) 0, (B) 10, (C) 12, (D) -12. This suggests a possible error in the question or the provided options.

Conclusion: Based on standard determinant calculation, the answer is 14, which is not listed.

The question asks to evaluate the truthfulness of an Assertion (A) and its Reason (R) regarding skew-symmetric matrices of odd order.

Assertion (A): If $A$ is a skew-symmetric matrix of odd order, then $|A|=0$.

Reason (R): For any skew-symmetric matrix $A$, $A' = -A$, so $|A'| = |-A|$. Also $|A'| = |A|$ and $|-A| = (-1)^n |A|$ for matrix of order $n$. If $n$ is odd, $|-A| = -|A|$. Thus $|A| = -|A|$, which implies $2|A|=0$, so $|A|=0$.

Analysis of Assertion (A):

A matrix $A$ is skew-symmetric if its transpose $A'$ is equal to its negative, i.e., $A' = -A$.

We know that the determinant of a transpose is equal to the determinant of the original matrix: $|A'| = |A|$.

We also know that for an $n \times n$ matrix $A$, $|kA| = k^n |A|$.

If $A$ is skew-symmetric, then $A' = -A$.

Taking the determinant of both sides:

Using the properties of determinants:

$|A| = (-1)^n |A|$

Now, consider the case where the order $n$ of the matrix $A$ is odd.

If $n$ is odd, then $(-1)^n = -1$.

So, the equation becomes:

$|A| = -|A|$

Rearranging this equation:

$|A| + |A| = 0$

$2|A| = 0$

Dividing by 2, we get:

$|A| = 0$

This proves that if $A$ is a skew-symmetric matrix of odd order, its determinant is 0.

Therefore, Assertion (A) is true.

Analysis of Reason (R):

The reason states: "For any skew-symmetric matrix $A$, $A' = -A$, so $|A'| = |-A|$. Also $|A'| = |A|$ and $|-A| = (-1)^n |A|$ for matrix of order $n$. If $n$ is odd, $|-A| = -|A|$. Thus $|A| = -|A|$, which implies $2|A|=0$, so $|A|=0$."

Let's break down the steps in the reason:

• "$A' = -A$": This is the definition of a skew-symmetric matrix, which is correct.
• "$|A'| = |-A|$": This follows directly from $A' = -A$ by taking the determinant of both sides, which is correct.
• "$|A'| = |A|$": This is a fundamental property of determinants, so it is correct.
• "$|-A| = (-1)^n |A|$ for matrix of order $n$": This is also a correct property of determinants.
• "If $n$ is odd, $|-A| = -|A|$": This correctly applies the property $|kA| = k^n |A|$ when $n$ is odd, making $(-1)^n = -1$.
• "Thus $|A| = -|A|$, which implies $2|A|=0$, so $|A|=0$": This correctly deduces that $|A|=0$ from $|A| = -|A|$.

The reason correctly explains the steps involved in proving that the determinant of a skew-symmetric matrix of odd order is zero.

Therefore, Reason (R) is true and it is the correct explanation for Assertion (A).

Conclusion:

Both Assertion (A) and Reason (R) are true, and Reason (R) provides a valid and correct explanation for Assertion (A).

Final Answer:

The correct option is (A) Both A and R are true and R is the correct explanation of A.

The question asks for the possible value of the determinant of a $3 \times 3$ matrix $A$, given its adjoint matrix.

Given:

• $A$ is a $3 \times 3$ matrix.
• $adj(A) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$, which is the identity matrix $I$.

To Find: The possible value(s) of $|A|$.

Options:

(A) $1$

(B) $-1$

(C) $\pm 1$

(D) $0$

Solution:

We know the relationship between a matrix, its adjoint, and its determinant:

Also, there is a property relating the determinant of the adjoint of a matrix to the determinant of the matrix itself:

where $n$ is the order of the square matrix.

In this problem, $n=3$. So, the property becomes:

$|adj(A)| = |A|^{3-1} = |A|^2$

We are given $adj(A) = I = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$.

Let's find the determinant of $adj(A)$:

$|adj(A)| = |I|$

The determinant of the identity matrix is always 1.

$|adj(A)| = 1$

Now, substitute this value into the property $|adj(A)| = |A|^2$:

$1 = |A|^2$

To find $|A|$, we take the square root of both sides:

$|A| = \pm \sqrt{1}$

$|A| = \pm 1$

This means the determinant of matrix $A$ can be either 1 or -1.

Final Answer:

The possible value(s) of $|A|$ are $\pm 1$.

The correct option is (C).

The question asks for the conditions on $a$ and $b$ that guarantee infinitely many solutions for a given system of linear equations, mentioning the determinant of the coefficient matrix and the condition $(adj A)B = O$.

Given System of Linear Equations:

1. $x + ay = 3$
2. $2x + by = 6$

In matrix form, this system can be written as $AX = B$, where:

$A = \begin{bmatrix} 1 & a \\ 2 & b \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 3 \\ 6 \end{bmatrix}$

Conditions for Infinitely Many Solutions:

A system of linear equations $AX=B$ has infinitely many solutions if:

1. The determinant of the coefficient matrix $A$ is zero ($|A|=0$).
2. The system is consistent. For a system with $|A|=0$, consistency implies that the augmented matrix $[A|B]$ has the same rank as $A$. This often translates to conditions involving the adjoint matrix or proportional relationships between the columns of $A$ and $B$.

Step 1: Calculate the determinant of the coefficient matrix $A$.

$|A| = (1 \times b) - (a \times 2)$

$|A| = b - 2a$

For infinitely many solutions, $|A|$ must be zero:

This implies $b = 2a$.

Step 2: Check the consistency condition using the given options.

We know that if $|A|=0$, the system can have either no solution or infinitely many solutions. For infinitely many solutions, the system must be consistent. The condition $(adj A)B = O$ is a way to check for consistency when $|A|=0$.

Let's evaluate the options based on $b=2a$ and the consistency criteria.

Option (A): $b = 2a$ and $B$ is proportional to one of the columns of $A$'s adjoint.

If $b=2a$, then $A = \begin{bmatrix} 1 & a \\ 2 & 2a \end{bmatrix}$.

$adj(A) = \begin{bmatrix} b & -a \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} 2a & -a \\ -2 & 1 \end{bmatrix}$ (for $b=2a$).

Let's check if $B$ is proportional to a column of $adj(A)$.

Column 1 of $adj(A)$ is $\begin{bmatrix} 2a \\ -2 \end{bmatrix}$. Is $\begin{bmatrix} 3 \\ 6 \end{bmatrix}$ proportional to $\begin{bmatrix} 2a \\ -2 \end{bmatrix}$?

This would mean $\frac{3}{2a} = \frac{6}{-2}$, so $3(-2) = 6(2a) \implies -6 = 12a \implies a = -1/2$. If $a=-1/2$, then $b=2a=-1$. This is a specific case.

Column 2 of $adj(A)$ is $\begin{bmatrix} -a \\ 1 \end{bmatrix}$. Is $\begin{bmatrix} 3 \\ 6 \end{bmatrix}$ proportional to $\begin{bmatrix} -a \\ 1 \end{bmatrix}$?

This would mean $\frac{3}{-a} = \frac{6}{1}$, so $3(1) = 6(-a) \implies 3 = -6a \implies a = -1/2$. Again, $a=-1/2$. This condition is not generally guaranteed just by $b=2a$. Option (A) seems too specific.

Option (B): $b = 2a$ and the matrix $\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$ has determinant zero.

The condition $b=2a$ ensures $|A|=0$.

The matrix $\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$ is formed by the coefficients of $x$ and the constant terms from the two equations. The determinant of this matrix is $(1 \times 6) - (3 \times 2) = 6 - 6 = 0$.

This condition (determinant of the matrix formed by the coefficients of $x$ and the constants being zero) is equivalent to checking if the second equation is a multiple of the first equation, which ensures consistency when $|A|=0$. If the second equation's coefficients and constant term are proportional to the first, and the proportionality constant is the same, then the equations are dependent, leading to infinitely many solutions.

Here, the ratio of coefficients of $x$ is $2/1=2$. The ratio of coefficients of $y$ is $b/a$. The ratio of constants is $6/3=2$. For infinitely many solutions, we need these ratios to be equal: $2/1 = b/a = 6/3$. This implies $b/a = 2$, so $b=2a$. And $6/3=2$ is already true. The condition that the determinant of $\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$ is zero is essentially checking if the ratio of constants to the first column coefficients is consistent, which is part of checking for consistency.

Option (C): $b=2a$ and the system is consistent.

This option correctly states that $b=2a$ is necessary for $|A|=0$. The condition "the system is consistent" is the second part required for infinitely many solutions when $|A|=0$. If $|A|=0$ and the system is consistent, then there are infinitely many solutions.

Let's verify consistency with $b=2a$. The system is $x+ay=3$ and $2x+2ay=6$. The second equation is $2(x+ay)=6$, which simplifies to $x+ay=3$. Both equations are identical, meaning the system is consistent and has infinitely many solutions.

Option (D): $b=2a$ and the points $(1,2)$ and $(a,b)$ are linearly dependent.

The condition $b=2a$ means that the vector $(a,b)$ is $(a, 2a)$. This vector is indeed linearly dependent with $(1,2)$ if $(a, 2a)$ is a multiple of $(1,2)$. This implies $(a, 2a) = k(1,2)$ for some scalar $k$. So $a=k$ and $2a=2k$. This means $a=k$. This condition is always met if $b=2a$ as the vector is $(a, 2a) = a(1, 2)$. The first column of matrix $A$ is $\begin{bmatrix} 1 \\ 2 \end{bmatrix}$. The second column is $\begin{bmatrix} a \\ b \end{bmatrix}$. Linear dependence of columns means one is a multiple of the other. If $b=2a$, then the second column is $\begin{bmatrix} a \\ 2a \end{bmatrix} = a \begin{bmatrix} 1 \\ 2 \end{bmatrix}$. This means the columns are linearly dependent if $b=2a$. This condition is equivalent to $|A|=0$. This option essentially restates $|A|=0$. It doesn't fully capture the consistency aspect needed for infinitely many solutions, though in this specific case, $b=2a$ implies consistency.

Comparing Options (B) and (C):

Option (B) states $b=2a$ and that a specific related matrix has a determinant of zero. This is a valid condition. The matrix $\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}$ represents the coefficients of $x$ and the constants. Its determinant being zero means the ratio of constants to the first column coefficients is consistent with the ratios of other coefficients.

Option (C) states $b=2a$ and that the system is consistent. This is the most direct and fundamental condition for infinitely many solutions when $|A|=0$. As shown earlier, if $b=2a$, the two equations become identical, thus the system is consistent and has infinitely many solutions.

Option (C) is the most encompassing and direct statement of the conditions required for infinitely many solutions.

Final Answer:

The conditions on $a$ and $b$ that guarantee infinitely many solutions are $b=2a$ (which makes the determinant of the coefficient matrix zero) and that the system is consistent. When $b=2a$, the second equation becomes a multiple of the first, ensuring consistency.

The correct option is (C).

The question asks to identify the statement that is NOT necessarily true if a matrix $A$ is invertible.

Definition of Invertible Matrix:

A matrix $A$ is invertible if there exists a matrix $A^{-1}$ such that $A A^{-1} = A^{-1} A = I$, where $I$ is the identity matrix.

Analysis of each option:

(A) $A$ is a square matrix.

For a matrix to have an inverse, it must be possible to multiply it by another matrix to get the identity matrix. The identity matrix is always a square matrix. For the product $AA^{-1}=I$ to hold, $A$ and $A^{-1}$ must have compatible dimensions. If $A$ is $m \times n$ and $A^{-1}$ is $p \times q$, then $np=mq$ and $pm=nq$. For $AA^{-1}=I_k$, we need $k \times k$ dimensions. For $A^{-1}A=I_m$, we need $m \times m$ dimensions. This implies that $m=n=k$. Therefore, an invertible matrix must be a square matrix.

This statement is necessarily true.

(B) $|A| \neq 0$.

A fundamental property of invertible matrices is that their determinant is non-zero. If $|A|=0$, the matrix is singular and does not have an inverse.

This statement is necessarily true.

(C) $A^{-1}$ exists and is unique.

The definition of an invertible matrix implies that its inverse $A^{-1}$ exists. Furthermore, the inverse of a matrix, if it exists, is unique. This can be proven by assuming there are two inverses and showing they must be equal.

This statement is necessarily true.

(D) $A$ is symmetric.

A matrix is symmetric if it is equal to its transpose ($A = A'$). While some invertible matrices are symmetric (e.g., the identity matrix), invertibility does not require a matrix to be symmetric. For example, consider the matrix $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$. Its determinant is $(1 \times 1) - (1 \times 0) = 1 \neq 0$, so it is invertible. However, its transpose is $A' = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$, which is not equal to $A$. Thus, $A$ is not symmetric.

This statement is NOT necessarily true.

Conclusion:

The statement that is NOT necessarily true if $A$ is an invertible matrix is that $A$ is symmetric.

Final Answer:

The correct option is (D).

The question asks to find the value of $|adj(A^2)|$ given that $A$ is a $3 \times 3$ matrix and $|A|=7$.

Given:

• $A$ is a $3 \times 3$ matrix.
• $|A| = 7$.

To Find: $|adj(A^2)|$.

Options:

(A) $7^2$

(B) $7^4$

(C) $7^6$

(D) $7^8$

Solution:

We will use two key properties of determinants and adjoints:

1. For an $n \times n$ matrix $B$, $|adj(B)| = |B|^{n-1}$.
2. For matrices $A$ and $B$ of the same order, $|AB| = |A||B|$.

Let $B = A^2$. We need to find $|adj(B)| = |adj(A^2)|$.

Using property 1, with $B = A^2$ and $n=3$, we have:

$|adj(A^2)| = |A^2|^2$

Now, we need to find $|A^2|$. Using property 2, with $A^2 = A \cdot A$, we have:

$|A^2| = |A \cdot A| = |A| \cdot |A| = |A|^2$

We are given $|A| = 7$. So:

$|A^2| = 7^2$

Substitute this value of $|A^2|$ back into equation (i):

$|adj(A^2)| = (|A^2|)^2 = (7^2)^2$

Using the exponent rule $(a^m)^n = a^{m \times n}$:

$|adj(A^2)| = 7^{2 \times 2} = 7^4$

Thus, $|adj(A^2)|$ is equal to $7^4$.

Final Answer:

The value of $|adj(A^2)|$ is $7^4$.

The correct option is (B).

The question asks to find the determinant of the matrix $(I-M)$, where $I$ is the identity matrix and $M$ is the given input-output matrix. The options provided represent steps or values involved in the calculation.

Given:

Input-output matrix $M = \begin{bmatrix} 0.2 & 0.5 \\ 0.4 & 0.1 \end{bmatrix}$.

Identity matrix $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

To Find: The determinant of the matrix $(I-M)$.

Options:

(A) $\begin{vmatrix} 0.8 & -0.5 \\ -0.4 & 0.9 \end{vmatrix}$

(B) $0.8 \times 0.9 - (-0.5) \times (-0.4)$

(C) $0.72 - 0.20 = 0.52$

(D) All of the above are steps/values involved.

Solution:

First, calculate the matrix $(I-M)$:

$I-M = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} - \begin{bmatrix} 0.2 & 0.5 \\ 0.4 & 0.1 \end{bmatrix}$

$I-M = \begin{bmatrix} 1 - 0.2 & 0 - 0.5 \\ 0 - 0.4 & 1 - 0.1 \end{bmatrix}$

$I-M = \begin{bmatrix} 0.8 & -0.5 \\ -0.4 & 0.9 \end{bmatrix}$

Now, let's analyze the options:

Option (A): $\begin{vmatrix} 0.8 & -0.5 \\ -0.4 & 0.9 \end{vmatrix}$

This option correctly represents the determinant of the matrix $(I-M)$ that we just calculated. This is the first step in finding the determinant.

Option (B): $0.8 \times 0.9 - (-0.5) \times (-0.4)$

This option shows the calculation of the determinant of the matrix $\begin{bmatrix} 0.8 & -0.5 \\ -0.4 & 0.9 \end{bmatrix}$ using the formula $ad - bc$. This is the second step in evaluating the determinant.

Option (C): $0.72 - 0.20 = 0.52$

This option shows the result of the calculation in Option (B). $0.8 \times 0.9 = 0.72$ and $(-0.5) \times (-0.4) = 0.20$. So, $0.72 - 0.20 = 0.52$. This is the final value of the determinant of $(I-M)$.

Option (D): All of the above are steps/values involved.

Since Options (A), (B), and (C) represent the matrix whose determinant is to be found, the calculation of its determinant, and the final value of the determinant respectively, all these are indeed steps or values involved in finding the determinant of $(I-M)$.

Conclusion:

All the options (A), (B), and (C) describe valid steps or values in the process of calculating the determinant of $(I-M)$. Therefore, option (D) is the most appropriate answer.

Final Answer:

The correct option is (D).

The question asks for the condition under which a system of homogeneous linear equations has a non-trivial solution.

Given System of Homogeneous Linear Equations:

1. $ax + by = 0$
2. $cx + dy = 0$

This system can be written in matrix form as $AX = O$, where:

$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, and $O = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$ (the zero vector).

Understanding Solutions of Homogeneous Systems:

A homogeneous system of linear equations always has at least one solution, which is the trivial solution ($x=0, y=0$).

The system has a non-trivial solution (i.e., at least one solution where $x$ or $y$ is not zero) if and only if the determinant of the coefficient matrix $A$ is zero.

Step 1: Calculate the determinant of the coefficient matrix $A$.

$|A| = ad - bc$

Step 2: Apply the condition for a non-trivial solution.

For the system to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

Analysis of Options:

• (A) $ad - bc \neq 0$: This condition ensures a unique trivial solution only.
• (B) $ad - bc = 0$: This is the condition for the existence of non-trivial solutions (infinitely many solutions).
• (C) $a=c$ and $b=d$: This condition would mean the two equations are $ax+by=0$ and $ax+by=0$, which implies infinitely many solutions, but it's a specific case of $ad-bc=0$ and not the general condition. For example, if $a=2, b=1, c=4, d=2$, then $ad-bc = (2)(2) - (1)(4) = 4-4=0$, leading to non-trivial solutions, but $a \neq c$ and $b \neq d$.
• (D) $a=b=c=d=0$: If all coefficients are zero, the equations become $0=0$ and $0=0$, which are true for all $x$ and $y$, so there are infinitely many non-trivial solutions. However, this is a trivial case where the matrix $A$ is the zero matrix, and its determinant is indeed 0. But the condition $ad-bc=0$ covers all cases where non-trivial solutions exist, including this one. The question asks for the general condition.

Final Answer:

The condition for the system of homogeneous linear equations to have a non-trivial solution is that the determinant of the coefficient matrix must be zero.

The correct option is (B).

The question asks us to find the value of $x$ given a determinant equation.

Given Equation:

$\begin{vmatrix} 2x+1 & 3 \\ x & 2 \end{vmatrix} = 5$

To Find: The value of $x$.

Options:

(A) $1$

(B) $2$

(C) $3$

(D) $4$

Solution:

The determinant of a $2 \times 2$ matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is given by $ad - bc$.

Applying this to the given equation:

$(2x+1)(2) - (3)(x) = 5$

Now, we solve this equation for $x$:

Distribute the 2:

$4x + 2 - 3x = 5$

Combine like terms ($4x$ and $-3x$):

$x + 2 = 5$

Subtract 2 from both sides of the equation:

$x = 5 - 2$

$x = 3$

So, the value of $x$ is 3.

Final Answer:

The value of $x$ is 3.

The correct option is (C).

The question asks for the adjoint of a given diagonal matrix $A$.

Given Matrix:

$A = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$

To Find: The adjoint of matrix $A$, $adj(A)$.

Options:

(A) $\begin{bmatrix} d & 0 \\ 0 & a \end{bmatrix}$

(B) $\begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$

(C) $\begin{bmatrix} -a & 0 \\ 0 & -d \end{bmatrix}$

(D) $\begin{bmatrix} d & 0 \\ 0 & a \end{bmatrix}$ if $ad \neq 0$, otherwise it doesn't exist.

Solution:

The adjoint of a matrix $A$, denoted as $adj(A)$, is the transpose of its cofactor matrix.

Let's find the cofactor matrix for $A = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$.

The elements of the matrix are $a_{11}=a$, $a_{12}=0$, $a_{21}=0$, $a_{22}=d$.

Cofactors:

• $C_{11} = (-1)^{1+1} M_{11} = (-1)^2 \times d = d$ (Minor $M_{11}$ is obtained by deleting row 1 and col 1, leaving $d$).
• $C_{12} = (-1)^{1+2} M_{12} = (-1)^3 \times 0 = 0$ (Minor $M_{12}$ is obtained by deleting row 1 and col 2, leaving $0$).
• $C_{21} = (-1)^{2+1} M_{21} = (-1)^3 \times 0 = 0$ (Minor $M_{21}$ is obtained by deleting row 2 and col 1, leaving $0$).
• $C_{22} = (-1)^{2+2} M_{22} = (-1)^4 \times a = a$ (Minor $M_{22}$ is obtained by deleting row 2 and col 2, leaving $a$).

The cofactor matrix is $\begin{bmatrix} C_{11} & C_{12} \\ C_{21} & C_{22} \end{bmatrix} = \begin{bmatrix} d & 0 \\ 0 & a \end{bmatrix}$.

The adjoint matrix is the transpose of the cofactor matrix:

$adj(A) = \begin{bmatrix} d & 0 \\ 0 & a \end{bmatrix}^T = \begin{bmatrix} d & 0 \\ 0 & a \end{bmatrix}$.

This result is valid regardless of whether $a$ or $d$ are zero, as long as the matrix $A$ itself exists. The determinant $|A| = ad - 0 = ad$. The inverse $A^{-1} = \frac{1}{ad} adj(A) = \frac{1}{ad} \begin{bmatrix} d & 0 \\ 0 & a \end{bmatrix}$. For the inverse to exist, we need $ad \neq 0$. However, the adjoint itself exists as long as the matrix A exists.

Let's examine the options:

• Option (A) matches our calculated adjoint.
• Option (B) is the original matrix $A$.
• Option (C) is $-A$.
• Option (D) states the adjoint is $\begin{bmatrix} d & 0 \\ 0 & a \end{bmatrix}$ if $ad \neq 0$, otherwise it doesn't exist. The adjoint exists even if $ad=0$. For example, if $A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, then $adj(A) = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}$, which is $\begin{bmatrix} d & 0 \\ 0 & a \end{bmatrix}$ with $d=0, a=1$. The adjoint exists.

Therefore, Option (A) correctly represents the adjoint of the given diagonal matrix.

Final Answer:

The adjoint of the matrix $A = \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix}$ is $\begin{bmatrix} d & 0 \\ 0 & a \end{bmatrix}$.

The correct option is (A).

The question asks about the property of the adjoint of a symmetric matrix.

Definition of Symmetric Matrix:

A matrix $A$ is symmetric if $A = A^T$. This means that the element at the $i$-th row and $j$-th column is equal to the element at the $j$-th row and $i$-th column ($a_{ij} = a_{ji}$).

Adjoint of a Matrix:

The adjoint of a matrix $A$, denoted as $adj(A)$, is the transpose of its cofactor matrix ($adj(A) = C^T$). The cofactor $C_{ij}$ of an element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor of $a_{ij}$.

Relationship between Symmetry of $A$ and $adj(A)$:

Let $A$ be a symmetric matrix, so $a_{ij} = a_{ji}$ for all $i, j$.

Let $C$ be the cofactor matrix of $A$. The element $C_{ij} = (-1)^{i+j} M_{ij}$.

The element $C_{ji} = (-1)^{j+i} M_{ji}$.

Since $A$ is symmetric, the submatrix obtained by deleting the $i$-th row and $j$-th column (to find $M_{ij}$) and the submatrix obtained by deleting the $j$-th row and $i$-th column (to find $M_{ji}$) are related.

Specifically, if $A$ is symmetric, then $M_{ij} = M_{ji}$. This is because the elements are mirrored across the main diagonal. To see this, consider forming the submatrix for $M_{ij}$ by deleting row $i$ and column $j$. The remaining elements are $a_{kl}$ where $k \neq i, l \neq j$. For $M_{ji}$, we delete row $j$ and column $i$. The remaining elements are $a_{kl}$ where $k \neq j, l \neq i$. Due to symmetry ($a_{kl} = a_{lk}$), the determinants of these submatrices are equal.

So, $M_{ij} = M_{ji}$.

Now let's look at the cofactors:

$C_{ij} = (-1)^{i+j} M_{ij}$

$C_{ji} = (-1)^{j+i} M_{ji}$

Since $i+j = j+i$, the terms $(-1)^{i+j}$ and $(-1)^{j+i}$ are the same.

Therefore, $C_{ij} = (-1)^{i+j} M_{ij}$ and $C_{ji} = (-1)^{i+j} M_{ij}$.

This implies $C_{ij} = C_{ji}$ for all $i, j$. Thus, the cofactor matrix $C$ is also symmetric.

The adjoint matrix is the transpose of the cofactor matrix: $adj(A) = C^T$.

Since $C$ is symmetric, $C^T = C$. Therefore, $adj(A) = C$.

This means that $adj(A)$ is equal to its transpose, so $adj(A)$ is also symmetric.

This holds true for any square matrix $A$ where $M_{ij} = M_{ji}$. For symmetric matrices, this is guaranteed.

Let's consider an example:

Let $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$. $A$ is symmetric.

$M_{11} = 3$, $M_{12} = 2$, $M_{21} = 2$, $M_{22} = 1$.

$C_{11} = 3$, $C_{12} = -2$, $C_{21} = -2$, $C_{22} = 1$.

Cofactor matrix $C = \begin{bmatrix} 3 & -2 \\ -2 & 1 \end{bmatrix}$, which is symmetric.

$adj(A) = C^T = \begin{bmatrix} 3 & -2 \\ -2 & 1 \end{bmatrix}$. $adj(A)$ is also symmetric.

This property holds true for any square matrix $A$ where $M_{ij} = M_{ji}$, which is true if $A$ is symmetric.

Analysis of Options:

(A) $adj(A)$ is always symmetric: Based on our derivation, this is correct.

(B) $adj(A)$ is always skew-symmetric: This is incorrect. A matrix is skew-symmetric if $adj(A) = -(adj(A))^T$.

(C) $adj(A)$ is diagonal: This is not always true. For example, $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$ is symmetric, but its adjoint is $\begin{bmatrix} 3 & -2 \\ -2 & 1 \end{bmatrix}$, which is not diagonal.

(D) $adj(A)$'s symmetry depends on the order of the matrix: While the existence of certain inverses or properties can depend on the order, the symmetry of the adjoint of a symmetric matrix does not depend on the order. Our proof was general for any $n \times n$ matrix where $M_{ij} = M_{ji}$.

Final Answer:

If $A$ is a symmetric matrix, its adjoint $adj(A)$ is always symmetric.

The correct option is (A).

The question asks about the effect of multiplying a row of a matrix by a scalar $k$ on its determinant.

Property of Determinants:

One of the fundamental properties of determinants states that if a matrix $B$ is obtained from a matrix $A$ by multiplying a single row (or a single column) by a scalar $k$, then the determinant of $B$ is $k$ times the determinant of $A$.

That is, if $B$ is obtained from $A$ by $R_i \rightarrow k R_i$, then $|B| = k |A|$.

Let's illustrate with a $2 \times 2$ matrix:

Let $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, so $|A| = ad - bc$.

If we multiply the first row by $k$ to get matrix $B$, then $B = \begin{bmatrix} ka & kb \\ c & d \end{bmatrix}$.

$|B| = (ka)d - (kb)c = k(ad - bc) = k|A|$.

If we multiply the second row by $k$ to get matrix $C$, then $C = \begin{bmatrix} a & b \\ kc & kd \end{bmatrix}$.

$|C| = a(kd) - b(kc) = k(ad - bc) = k|A|$.

Note: Option (C) $k^n$ applies when the entire matrix is multiplied by $k$ (i.e., $kA$), not just a single row.

Analysis of Options:

(A) $k$: This matches the property that multiplying a single row by $k$ multiplies the determinant by $k$.

(B) $1/k$: This would be the case if we were dividing a row by $k$.

(C) $k^n$ (where $n$ is the order): This is the property for $|kA|$. If $A$ is $n \times n$, then $kA$ has all $n$ rows multiplied by $k$, so the determinant is multiplied by $k^n$. This is not the case here where only one row is multiplied.

(D) $1$: This would mean the determinant remains unchanged, which is true for row operations like $R_i \rightarrow R_i + k R_j$, but not for multiplying a row by a scalar.

Final Answer:

The determinant of a matrix obtained by multiplying a row by a scalar $k$ is $k$ times the original determinant.

The correct option is (A).

The question asks to find the determinant of the adjoint of an invertible matrix $A$, given its order and determinant.

Given:

• $A$ is an invertible matrix of order 2 ($n=2$).
• $|A| = 3$.

To Find: $|adj(A)|$.

Options:

(A) $3$

(B) $9$

(C) $1/3$

(D) $1/9$

Solution:

We use the property relating the determinant of the adjoint of a matrix to the determinant of the matrix itself:

where $n$ is the order of the square matrix $A$.

In this problem, we are given:

• $n = 2$ (order of the matrix)
• $|A| = 3$ (determinant of matrix A)

Substitute these values into the formula (i):

$|adj(A)| = |A|^{2-1}$

$|adj(A)| = |A|^1$

$|adj(A)| = |A|$

Since $|A| = 3$, we have:

$|adj(A)| = 3$

Final Answer:

The value of $|adj(A)|$ is 3.

The correct option is (A).

The question asks to find the value of a given 3x3 determinant.

Given Determinant:

$D = \begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix}$

To Find: The value of the determinant $D$.

Options:

(A) $a+b+c$

(B) $(a+b+c)(a-b)(b-c)(c-a)$

(C) $0$

(D) $abc$

Solution:

We can simplify the determinant using row and column operations.

Let $C_1, C_2, C_3$ denote the columns of the determinant.

Apply the column operation $C_3 \rightarrow C_3 + C_2$: This operation adds the elements of the second column to the elements of the third column.

$D = \begin{vmatrix} 1 & a & (b+c)+a \\ 1 & b & (c+a)+b \\ 1 & c & (a+b)+c \end{vmatrix}$

$D = \begin{vmatrix} 1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c \end{vmatrix}$

Now, we can factor out $(a+b+c)$ from the third column:

$D = (a+b+c) \begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix}$

Observe the resulting determinant $\begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix}$. This determinant has two identical columns (the first and the third columns are both filled with 1s).

A property of determinants states that if a matrix has two identical columns (or two identical rows), its determinant is zero.

Therefore, $\begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix} = 0$.

Substituting this back into the expression for $D$:

$D = (a+b+c) \times 0$

$D = 0$

Thus, the value of the determinant is 0.

Final Answer:

The value of the determinant is 0.

The correct option is (C).

The question provides a matrix equation $A^2 - 3A + 2I = O$ and asks for a possible expression for the inverse of matrix $A$, $A^{-1}$.

Given Equation:

$A^2 - 3A + 2I = O$

where $A$ is a square matrix, $I$ is the identity matrix, and $O$ is the zero matrix.

To Find: A possible expression for $A^{-1}$.

Options:

(A) $3I - A$

(B) $3I - \frac{1}{2}A$

(C) $\frac{1}{2}(3I - A)$

(D) $A - 3I$

Solution:

To find the inverse $A^{-1}$, we need to manipulate the given equation such that one side contains $A^{-1}$ multiplied by a matrix, and the other side is a combination of $I$ and $A$. We can do this by multiplying the entire equation by $A^{-1}$ (assuming $A$ is invertible).

First, rearrange the given equation to isolate the term with $A^{-1}$:

$A^2 - 3A = -2I$

Factor out $A$ from the left side:

$A(A - 3I) = -2I$

Now, multiply both sides by $A^{-1}$ on the left:

$A^{-1} [A(A - 3I)] = A^{-1} (-2I)$

Using the associative property of matrix multiplication $(A^{-1}A)(A-3I) = -2A^{-1}$:

$I(A - 3I) = -2A^{-1}$

Since $I$ times any matrix is the matrix itself ($I X = X$):

$A - 3I = -2A^{-1}$

Now, we need to solve for $A^{-1}$. Divide both sides by -2:

$A^{-1} = \frac{A - 3I}{-2}$

$A^{-1} = -\frac{1}{2} (A - 3I)$

Distribute the $-\frac{1}{2}$:

$A^{-1} = -\frac{1}{2}A + \frac{3}{2}I$

$A^{-1} = \frac{3}{2}I - \frac{1}{2}A$

We can rewrite this by factoring out $\frac{1}{2}$:

$A^{-1} = \frac{1}{2}(3I - A)$

Let's compare this result with the given options:

• Option (A) is $3I - A$. This is $-2A^{-1}$.
• Option (B) is $3I - \frac{1}{2}A$. This is not equal to our result.
• Option (C) is $\frac{1}{2}(3I - A)$. This matches our derived expression for $A^{-1}$.
• Option (D) is $A - 3I$. This is equal to $-2A^{-1}$.

Final Answer:

The possible expression for $A^{-1}$ is $\frac{1}{2}(3I - A)$.

The correct option is (C).

The question asks to find the value of $x$ for a given system of linear equations using Cramer's rule.

Given System of Linear Equations:

1. $2x - y = 5$
2. $x + 3y = -1$

To Find: The value of $x$ using Cramer's rule.

Options:

(A) $1$

(B) $2$

(C) $3$

(D) $-1$

Solution:

Cramer's rule states that for a system of linear equations $AX = B$, the solution for a variable $x$ is given by $x = \frac{|A_x|}{|A|}$, where:

• $|A|$ is the determinant of the coefficient matrix $A$.
• $|A_x|$ is the determinant of the matrix formed by replacing the first column (coefficients of $x$) of $A$ with the constant terms $B$.

First, let's write the system in matrix form $AX = B$:

$A = \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \end{bmatrix}$, $B = \begin{bmatrix} 5 \\ -1 \end{bmatrix}$

Step 1: Calculate the determinant of the coefficient matrix $A$.

$|A| = \begin{vmatrix} 2 & -1 \\ 1 & 3 \end{vmatrix}$

$|A| = (2 \times 3) - (-1 \times 1)$

$|A| = 6 - (-1)$

$|A| = 6 + 1 = 7$

Step 2: Calculate the determinant $|A_x|$.

To find $|A_x|$, we replace the first column of $A$ (the coefficients of $x$) with the constant terms from $B$:

$A_x = \begin{bmatrix} 5 & -1 \\ -1 & 3 \end{bmatrix}$

$|A_x| = \begin{vmatrix} 5 & -1 \\ -1 & 3 \end{vmatrix}$

$|A_x| = (5 \times 3) - (-1 \times -1)$

$|A_x| = 15 - 1$

$|A_x| = 14$

Step 3: Apply Cramer's rule to find $x$.

$x = \frac{|A_x|}{|A|}$

$x = \frac{14}{7}$

$x = 2$

The value of $x$ is 2.

Final Answer:

The value of $x$ is 2.

The correct option is (B).

To find the values of $p$ for which the points $(p+1, 1)$, $(2p+1, 3)$, and $(2p+2, 2p)$ are collinear, we can use the slope condition: the slope between the first two points must equal the slope between the second two points.

Let the points be $A=(p+1, 1)$, $B=(2p+1, 3)$, and $C=(2p+2, 2p)$.

The slope of AB is $m_{AB} = \frac{3-1}{(2p+1)-(p+1)} = \frac{2}{p}$.

The slope of BC is $m_{BC} = \frac{2p-3}{(2p+2)-(2p+1)} = \frac{2p-3}{1} = 2p-3$.

For collinearity, $m_{AB} = m_{BC}$ (assuming $p \neq 0$):

$\frac{2}{p} = 2p-3$

$2 = p(2p-3)$

$2 = 2p^2 - 3p$

$2p^2 - 3p - 2 = 0$

Factor the quadratic equation:

$(2p+1)(p-2) = 0$

This gives the solutions:

$2p+1 = 0 \implies p = -1/2$

$p-2 = 0 \implies p = 2$

The values of $p$ are $-1/2$ and $2$.

Checking the options:

My derived solutions are $p=-1/2$ and $p=2$. None of the provided options (A) $-1$ or $1/2$, (B) $1$ or $-1/2$, (C) $0$ or $1$, (D) $-1$ or $-1/2$ match this pair of solutions.

However, $p=-1/2$ is a correct solution and appears in options (A), (B), and (D). Given this discrepancy, there is likely an error in the question or the provided options.

Conclusion based on discrepancy:

As the derived correct solutions are $p=-1/2$ and $p=2$, and this pair is not available in the options, a definitive answer from the given choices cannot be provided. However, $p=-1/2$ is confirmed to be a correct value.

The question asks to find the cofactor of the element $a_{22}$ in the given 2x2 matrix $A$.

Given Matrix:

$A = \begin{bmatrix} 2 & 3 \\ -1 & 4 \end{bmatrix}$

To Find: The cofactor of the element $a_{22}$.

Options:

(A) $-1$

(B) $4$

(C) $2$

(D) $3$

Solution:

The element $a_{22}$ is the element in the 2nd row and 2nd column of matrix $A$. In the given matrix, $a_{22} = 4$.

The cofactor of an element $a_{ij}$ is given by the formula $C_{ij} = (-1)^{i+j}M_{ij}$, where $M_{ij}$ is the minor of the element $a_{ij}$.

The minor $M_{ij}$ is the determinant of the submatrix obtained by deleting the $i$-th row and the $j$-th column of the original matrix.

For the element $a_{22}$, we need to find $C_{22}$. Here, $i=2$ and $j=2$.

First, let's find the minor $M_{22}$ by deleting the 2nd row and 2nd column of matrix A:

$A = \begin{bmatrix} 2 & \cancel{3} \\ \cancel{-1} & \cancel{4} \end{bmatrix}$

The submatrix consists of the single element $2$. The minor $M_{22}$ is the determinant of this submatrix, which is just the element itself.

$M_{22} = 2$

Now, we can find the cofactor $C_{22}$ using the formula $C_{ij} = (-1)^{i+j}M_{ij}$:

$C_{22} = (-1)^{2+2}M_{22}$

$C_{22} = (-1)^{4} \times 2$

$C_{22} = 1 \times 2$

$C_{22} = 2$

Thus, the cofactor of the element $a_{22}$ is 2.

Final Answer:

The cofactor of the element $a_{22}$ is 2.

The correct option is (C).

The question asks to evaluate the determinant of a given 2x2 matrix.

Given Determinant:

$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$

To Evaluate: The value of the determinant.

Solution:

The determinant of a 2x2 matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is calculated using the formula $ad - bc$.

In this case, we have:

• $a = 2$
• $b = 4$
• $c = -5$
• $d = -1$

Applying the formula:

Now, perform the multiplication and subtraction:

$= -2 - (-20)$

$= -2 + 20$

$= 18$

Therefore, the value of the determinant is 18.

Final Answer:

The value of the determinant $\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$ is 18.

The question asks to find the value(s) of $x$ by equating two determinants.

Given Equation:

$\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$

To Find: The value(s) of $x$.

Solution:

First, we evaluate the determinant on the left side of the equation:

Left side determinant: $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = (x)(x) - (2)(18) = x^2 - 36$.

Next, we evaluate the determinant on the right side of the equation:

Right side determinant: $\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix} = (6)(6) - (2)(18) = 36 - 36 = 0$.

Now, we set the two determinants equal to each other:

$x^2 - 36 = 0$

To solve for $x$, we can add 36 to both sides:

$x^2 = 36$

Take the square root of both sides:

$x = \pm \sqrt{36}$

$x = \pm 6$

So, the possible values of $x$ are $6$ and $-6$.

Final Answer:

The values of $x$ are $6$ and $-6$.

The question asks to evaluate the determinant of a given 3x3 matrix.

Given Determinant:

$D = \begin{vmatrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{vmatrix}$

To Evaluate: The value of the determinant $D$.

Solution:

We can evaluate the determinant by expanding along any row or column. Expanding along the third column is convenient because it contains two zeros.

The formula for expansion along the third column ($C_3$) is:

where $a_{ij}$ is the element in the $i$-th row and $j$-th column, and $C_{ij}$ is its cofactor.

The elements in the third column are $a_{13}=4$, $a_{23}=0$, and $a_{33}=0$.

So, the formula simplifies to:

$D = 4 \times C_{13} + 0 \times C_{23} + 0 \times C_{33}$

$D = 4 \times C_{13}$

Now, we need to find the cofactor $C_{13}$. The cofactor $C_{ij} = (-1)^{i+j} M_{ij}$, where $M_{ij}$ is the minor (determinant of the submatrix).

For $C_{13}$ ($i=1, j=3$):

$C_{13} = (-1)^{1+3} M_{13}$

$C_{13} = (-1)^{4} M_{13}$

$C_{13} = 1 \times M_{13}$

The minor $M_{13}$ is found by deleting the 1st row and 3rd column of the original matrix:

$M_{13} = \begin{vmatrix} -1 & 3 \\ 4 & 1 \end{vmatrix}$

Calculate the 2x2 determinant:

$M_{13} = (-1)(1) - (3)(4)$

$M_{13} = -1 - 12$

$M_{13} = -13$

Now, substitute $M_{13}$ back into the cofactor formula:

$C_{13} = 1 \times (-13) = -13$

Finally, substitute $C_{13}$ back into the determinant formula for $D$:

$D = 4 \times C_{13} = 4 \times (-13)$

$D = -52$

Therefore, the value of the determinant is -52.

Final Answer:

The value of the determinant is -52.

The question asks to find the area of a triangle given the coordinates of its vertices.

Given Vertices:

• $(x_1, y_1) = (3, 8)$
• $(x_2, y_2) = (-4, 2)$
• $(x_3, y_3) = (5, 1)$

To Find: The area of the triangle formed by these vertices.

Solution:

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ can be calculated using the determinant formula:

Alternatively, using a determinant form:

Let's use the first formula (expansion form):

Substitute the coordinates into equation (i):

$Area = \frac{1}{2} \left| 3(2 - 1) + (-4)(1 - 8) + 5(8 - 2) \right|$

Perform the subtractions inside the parentheses:

$Area = \frac{1}{2} \left| 3(1) + (-4)(-7) + 5(6) \right|$

Perform the multiplications:

$Area = \frac{1}{2} \left| 3 + 28 + 30 \right|$

Add the numbers inside the absolute value:

$Area = \frac{1}{2} \left| 61 \right|$

$Area = \frac{1}{2} \times 61$

$Area = \frac{61}{2}$

The area is $\frac{61}{2}$ square units, or $30.5$ square units.

Final Answer:

The area of the triangle is $\frac{61}{2}$ square units.

The question asks to find the value of $k$ given that three points are collinear, using the determinant method.

Given Points:

• $(x_1, y_1) = (2, -3)$
• $(x_2, y_2) = (k, -1)$
• $(x_3, y_3) = (0, 4)$

Condition for Collinearity:

Three points are collinear if the area of the triangle formed by them is zero. Using the determinant method, this means:

Solution:

Substitute the given coordinates into equation (i):

Now, we evaluate the determinant. Expanding along the first column is convenient because of the zero element.

The expansion along the first column is $a_{11}C_{11} + a_{21}C_{21} + a_{31}C_{31}$:

$2 \times C_{11} + k \times C_{21} + 0 \times C_{31} = 0$

$2 \times C_{11} + k \times C_{21} = 0$

Calculate the cofactors:

For $C_{11}$: $(-1)^{1+1} M_{11} = 1 \times \begin{vmatrix} -1 & 1 \\ 4 & 1 \end{vmatrix} = (-1)(1) - (1)(4) = -1 - 4 = -5$.

For $C_{21}$: $(-1)^{2+1} M_{21} = (-1)^3 \times \begin{vmatrix} -3 & 1 \\ 4 & 1 \end{vmatrix} = -1 \times [(-3)(1) - (1)(4)] = -1 \times [-3 - 4] = -1 \times [-7] = 7$.

Substitute the cofactor values back into the equation:

$2 \times (-5) + k \times (7) = 0$

$-10 + 7k = 0$

Now, solve for $k$:

$7k = 10$

$k = \frac{10}{7}$

Alternatively, expanding along the first row:

$2 \begin{vmatrix} -1 & 1 \\ 4 & 1 \end{vmatrix} - (-3) \begin{vmatrix} k & 1 \\ 0 & 1 \end{vmatrix} + 1 \begin{vmatrix} k & -1 \\ 0 & 4 \end{vmatrix} = 0$

$2[(-1)(1) - (1)(4)] + 3[(k)(1) - (1)(0)] + 1[(k)(4) - (-1)(0)] = 0$

$2[-1 - 4] + 3[k - 0] + 1[4k - 0] = 0$

$2[-5] + 3[k] + 1[4k] = 0$

$-10 + 3k + 4k = 0$

$-10 + 7k = 0$

$7k = 10$

$k = \frac{10}{7}$

Both methods yield the same result.

Final Answer:

The value of $k$ is $\frac{10}{7}$.

The question asks to find the minor and cofactor of the element $a_{21}$ in the given 3x3 determinant.

Given Determinant:

$D = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$

To Find: The minor and cofactor of the element $a_{21}$.

Solution:

The element $a_{21}$ is the element in the 2nd row and 1st column of the determinant. In the given determinant, $a_{21} = 4$.

1. Minor of $a_{21}$ ($M_{21}$):

The minor of an element $a_{ij}$ is the determinant of the submatrix obtained by deleting the $i$-th row and the $j$-th column.

To find $M_{21}$, we delete the 2nd row and the 1st column of the determinant:

$D = \begin{vmatrix} \cancel{1} & 2 & 3 \\ \cancel{4} & \cancel{5} & \cancel{6} \\ \cancel{7} & 8 & 9 \end{vmatrix}$

The remaining elements form the submatrix:

$M_{21} = \begin{vmatrix} 2 & 3 \\ 8 & 9 \end{vmatrix}$

Now, we calculate the determinant of this 2x2 matrix:

$M_{21} = (2 \times 9) - (3 \times 8)$

$M_{21} = 18 - 24$

$M_{21} = -6$

So, the minor of $a_{21}$ is $-6$.

2. Cofactor of $a_{21}$ ($C_{21}$):

The cofactor of an element $a_{ij}$ is given by $C_{ij} = (-1)^{i+j} M_{ij}$.

For $a_{21}$, we have $i=2$ and $j=1$.

$C_{21} = (-1)^{2+1} M_{21}$

$C_{21} = (-1)^{3} \times (-6)$

$C_{21} = -1 \times (-6)$

$C_{21} = 6$

So, the cofactor of $a_{21}$ is $6$.

Final Answer:

The minor of the element $a_{21}$ is $-6$.

The cofactor of the element $a_{21}$ is $6$.

The question asks to find the adjoint of the given 2x2 matrix $A$.

Given Matrix:

$A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$

To Find: The adjoint of matrix $A$, denoted as $\text{adj}(A)$.

Solution:

The adjoint of a 2x2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by $\text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$. This involves swapping the diagonal elements and negating the off-diagonal elements.

For the given matrix $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$, we have:

• $a = 2$
• $b = 3$
• $c = 1$
• $d = 4$

Applying the formula for the adjoint:

Swap the diagonal elements ($a$ and $d$): $4$ and $2$.

Negate the off-diagonal elements ($b$ and $c$): $-3$ and $-1$.

So, the adjoint of $A$ is:

$\text{adj}(A) = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$

Final Answer:

The adjoint of the matrix $A$ is $\begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$.

The question asks to find the inverse of the given 2x2 matrix $A$.

Given Matrix:

$A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$

To Find: The inverse of matrix $A$, denoted as $A^{-1}$.

Solution:

The inverse of a 2x2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is given by the formula:

where $|A|$ is the determinant of $A$, and $\text{adj}(A)$ is the adjoint of $A$.

Step 1: Calculate the determinant of $A$.

$|A| = (2 \times 4) - (3 \times 1)$

$|A| = 8 - 3$

$|A| = 5$

Since $|A| \neq 0$, the inverse exists.

Step 2: Find the adjoint of $A$.

For a 2x2 matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the adjoint is $\text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

For $A = \begin{pmatrix} 2 & 3 \\ 1 & 4 \end{pmatrix}$:

Swap diagonal elements: $4$ and $2$.

Negate off-diagonal elements: $-3$ and $-1$.

So, $\text{adj}(A) = \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$.

Step 3: Calculate $A^{-1}$ using the formula (i).

$A^{-1} = \frac{1}{5} \begin{pmatrix} 4 & -3 \\ -1 & 2 \end{pmatrix}$

$A^{-1} = \begin{pmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{pmatrix}$

Final Answer:

The inverse of the matrix $A$ is $\begin{pmatrix} 4/5 & -3/5 \\ -1/5 & 2/5 \end{pmatrix}$.

The question asks to prove the property $|AB| = |A| |B|$ for 2x2 matrices $A$ and $B$.

To Prove: $|AB| = |A| |B|$ for $2 \times 2$ matrices $A$ and $B$.

Let:

Matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$

Matrix $B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$

Step 1: Calculate the determinants of $A$ and $B$.

$|A| = ad - bc$

$|B| = eh - fg$

Therefore, $|A||B| = (ad - bc)(eh - fg)$.

Step 2: Calculate the product matrix $AB$.

$AB = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} e & f \\ g & h \end{pmatrix}$

$AB = \begin{pmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix}$

Step 3: Calculate the determinant of $AB$.

$|AB| = \det \begin{pmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix}$

$|AB| = (ae+bg)(cf+dh) - (af+bh)(ce+dg)$

Expand the terms:

Term 1: $(ae+bg)(cf+dh) = aecf + aedh + bgcf + bgdh$

Term 2: $(af+bh)(ce+dg) = afce + afdg + bhce + bhdg$

Now subtract Term 2 from Term 1:

$|AB| = (aecf + aedh + bgcf + bgdh) - (afce + afdg + bhce + bhdg)$

$|AB| = aecf + aedh + bgcf + bgdh - afce - afdg - bhce - bhdg$

Notice that $aecf$ and $afce$ cancel out.

$|AB| = aedh + bgcf + bgdh - afdg - bhce - bhdg$

Let's regroup the terms to match $|A||B| = (ad - bc)(eh - fg) = adeh - adfg - bceh + bcfg$.

My expansion of $|AB|$ does not immediately match. Let me re-evaluate the expansion of $|AB|$ and try to factor it.

$|AB| = (ae+bg)(cf+dh) - (af+bh)(ce+dg)$

$= (aecf + aedh + bgcf + bgdh) - (afce + afdg + bhce + bhdg)$

$= aecf + aedh + bgcf + bgdh - afce - afdg - bhce - bhdg$

The terms $aecf$ and $afce$ are the same, so they cancel. Similarly $bgdh$ and $bhdg$ cancel.

$|AB| = aedh + bgcf - afdg - bhce$

Let's compare this with $|A||B| = adeh - adfg - bceh + bcfg$.

My expanded $|AB|$ is: $aedh + bgcf - afdg - bhce$.

Let's rewrite $|A||B|$ as: $adeh - adfg - bceh + bcfg$.

It seems my expansion of $|AB|$ is incorrect or needs significant rearrangement.

Let's retry the expansion of $|AB|$ and aim to factor it.

$|AB| = (ae+bg)(cf+dh) - (af+bh)(ce+dg)$

$= (a)(e)(c)(f) + (a)(e)(d)(h) + (b)(g)(c)(f) + (b)(g)(d)(h) - [(a)(f)(c)(e) + (a)(f)(d)(g) + (b)(h)(c)(e) + (b)(h)(d)(g)]$

$= aecf + aedh + bgcf + bgdh - afce - afdg - bhce - bhdg$

$= aedh + bgcf - afdg - bhce$ (after cancelling $aecf$ and $bgdh$)

Let's rearrange the terms of $|A||B| = (ad - bc)(eh - fg)$:

$|A||B| = adeh - adfg - bceh + bcfg$.

The terms in my $|AB|$ expansion ($aedh + bgcf - afdg - bhce$) do not directly match $adeh - adfg - bceh + bcfg$. There must be a mistake in my expansion or simplification of $|AB|$.

Let's use a property that states $|AB| = |A| |B|$. To prove this, we can use the fact that the determinant is a multilinear function of its rows (or columns). However, a direct algebraic proof is required here.

Let's retry the expansion of $|AB|$ carefully:

$AB = \begin{pmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix}$

$|AB| = (ae+bg)(cf+dh) - (af+bh)(ce+dg)$

$= (ae)(cf) + (ae)(dh) + (bg)(cf) + (bg)(dh) - [ (af)(ce) + (af)(dg) + (bh)(ce) + (bh)(dg) ]$

$= aecf + aedh + bgcf + bgdh - afce - afdg - bhce - bhdg$

$= aedh + bgcf - afdg - bhce$ (after canceling $aecf$ and $bgdh$).

Let's try to factor this expression to match $|A||B|$.

$aedh + bgcf - afdg - bhce$

Group terms related to $ad$ and $bc$:

$ad(eh - fg) - bc(fh - eg)$ --- this is not working directly.

Let's try grouping terms related to $a,b,c,d$ and $e,f,g,h$.

$|AB| = a(edh - fdg) + b(gcf - hce)$ --- Still not matching.

Let's go back to the original expansion of $|AB|$:

$|AB| = (ae+bg)(cf+dh) - (af+bh)(ce+dg)$

Let's group terms from $|A||B| = (ad - bc)(eh - fg) = adeh - adfg - bceh + bcfg$.

Consider the expression for $|AB|$ again:

$aedh + bgcf - afdg - bhce$

Rearrange to match $|A||B|$'s structure.

$adeh - adfg - bceh + bcfg$

Let's try factoring $|AB|$ differently:

$|AB| = aedh + bgcf - afdg - bhce$

$= ad(eh) - ad(fg) - bc(eh) + bc(fg)$ --- this is $|A||B|$

Let's group the terms in $|AB|$ as follows:

$|AB| = aedh - afdg + bgcf - bhce$

$= ad(eh - fg) + bc(gf - he)$

$= ad(eh - fg) - bc(he - gf)$

$= ad(eh - fg) - bc(eh - fg)$

$= (ad - bc)(eh - fg)$

$= |A||B|$

The factoring works. Let's verify the grouping:

$aedh - afdg + bgcf - bhce$

Group the first two terms: $ad(eh - fg)$

Group the last two terms: $bc(gf - he)$ which is $bc(-(he-gf)) = -bc(he-gf)$

So, $ad(eh - fg) - bc(he - gf)$. This matches $|A||B|$.

Thus, the algebraic expansion and factorization confirm that $|AB| = |A||B|$.

Conclusion:

We have shown through algebraic expansion and factorization that the determinant of the product of two 2x2 matrices is equal to the product of their individual determinants.

Proof Steps:

1. Define two general 2x2 matrices $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ and $B = \begin{pmatrix} e & f \\ g & h \end{pmatrix}$.

2. Calculate $|A| = ad-bc$ and $|B| = eh-fg$. Compute $|A||B| = (ad-bc)(eh-fg)$.

3. Calculate the product matrix $AB = \begin{pmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{pmatrix}$.

4. Calculate $|AB| = (ae+bg)(cf+dh) - (af+bh)(ce+dg)$.

5. Expand both expressions and show that they are equal after simplification and factorization.

Expansion of $|AB|$:

$|AB| = aedh + bgcf - afdg - bhce$

$= ad(eh - fg) + bc(gf - he)$

$= ad(eh - fg) - bc(he - gf)$

$= ad(eh - fg) - bc(eh - fg)$

$= (ad - bc)(eh - fg)$

$= |A||B|$

This completes the proof.

The question asks to evaluate the determinant of a given 3x3 matrix without expanding it directly, implying the use of determinant properties.

Given Determinant:

$D = \begin{vmatrix} 2 & 3 & 5 \\ 2 & 3 & 5 \\ 1 & 2 & 3 \end{vmatrix}$

To Evaluate: The value of the determinant $D$ without expansion.

Solution:

We observe that the first row and the second row of the determinant are identical. That is, $R_1 = R_2$.

A fundamental property of determinants states that if any two rows (or any two columns) of a determinant are identical, then the value of the determinant is zero.

Since the first row $\begin{pmatrix} 2 & 3 & 5 \end{pmatrix}$ is identical to the second row $\begin{pmatrix} 2 & 3 & 5 \end{pmatrix}$, the determinant must be zero.

Therefore, $\begin{vmatrix} 2 & 3 & 5 \\ 2 & 3 & 5 \\ 1 & 2 & 3 \end{vmatrix} = 0$.

Final Answer:

The value of the determinant is 0.

The question asks to prove that a given determinant is equal to zero using properties of determinants.

Given Determinant:

$D = \begin{vmatrix} a & b & c \\ a+x & b+x & c+x \\ x & x & x \end{vmatrix}$

To Prove: $D = 0$ using determinant properties.

Solution:

We can use row or column operations to simplify the determinant.

Let $R_1, R_2, R_3$ denote the rows of the determinant.

Consider the row operation $R_2 \rightarrow R_2 - R_1$. This operation subtracts the elements of the first row from the corresponding elements of the second row. This operation does not change the value of the determinant.

$D = \begin{vmatrix} a & b & c \\ (a+x)-a & (b+x)-b & (c+x)-c \\ x & x & x \end{vmatrix}$

$D = \begin{vmatrix} a & b & c \\ x & x & x \\ x & x & x \end{vmatrix}$

Now, we observe that the second row ($R_2$) and the third row ($R_3$) of the determinant are identical. That is, $R_2 = R_3$.

A property of determinants states that if any two rows (or any two columns) of a determinant are identical, then the value of the determinant is zero.

Since the second and third rows are identical, the determinant $D$ is zero.

$D = 0$

This proves that $\left| \begin{matrix} a & b & c \\ a+x & b+x & c+x \\ x & x & x \end{matrix} \right| = 0$.

Final Answer:

The determinant is proven to be 0 due to the presence of two identical rows after performing the row operation $R_2 \rightarrow R_2 - R_1$.

The question asks to find the value of $x$ for which the given matrix $A$ is singular.

Given Matrix:

$A = \begin{pmatrix} 3-x & 2 \\ 1 & 8-x \end{pmatrix}$

Condition for a Singular Matrix:

A matrix is singular if and only if its determinant is equal to zero.

Solution:

First, calculate the determinant of the matrix $A$. For a 2x2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the determinant is $ad - bc$.

For matrix $A = \begin{pmatrix} 3-x & 2 \\ 1 & 8-x \end{pmatrix}$:

$\det(A) = (3-x)(8-x) - (2)(1)$

Now, apply the condition for a singular matrix (equation (i)):

$(3-x)(8-x) - 2 = 0$

Expand the product $(3-x)(8-x)$:

$24 - 3x - 8x + x^2 - 2 = 0$

Combine like terms:

$x^2 - 11x + 24 - 2 = 0$

$x^2 - 11x + 22 = 0$

This is a quadratic equation in the form $ax^2 + bx + c = 0$, where $a=1$, $b=-11$, and $c=22$. We can solve for $x$ using the quadratic formula:

Substitute the values of $a, b, c$ into the formula:

$x = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(22)}}{2(1)}$

$x = \frac{11 \pm \sqrt{121 - 88}}{2}$

$x = \frac{11 \pm \sqrt{33}}{2}$

The values of $x$ for which the matrix $A$ is singular are $x = \frac{11 + \sqrt{33}}{2}$ and $x = \frac{11 - \sqrt{33}}{2}$.

Final Answer:

The values of $x$ for which the matrix $A$ is singular are $\frac{11 + \sqrt{33}}{2}$ and $\frac{11 - \sqrt{33}}{2}$.

The question asks to verify the property $\text{adj}(AB) = \text{adj}(B) \text{adj}(A)$ for the given matrices $A$ and $B$.

Given Matrices:

$A = \begin{pmatrix} 2 & 3 \\ 1 & -4 \end{pmatrix}$

$B = \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix}$

To Verify: $\text{adj}(AB) = \text{adj}(B) \text{adj}(A)$.

Solution:

We need to calculate both sides of the equation separately and then compare them.

Left Side: $\text{adj}(AB)$

Step 1: Calculate the product matrix $AB$.

$AB = \begin{pmatrix} 2 & 3 \\ 1 & -4 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix}$

$AB = \begin{pmatrix} (2)(1)+(3)(-1) & (2)(-2)+(3)(3) \\ (1)(1)+(-4)(-1) & (1)(-2)+(-4)(3) \end{pmatrix}$

$AB = \begin{pmatrix} 2-3 & -4+9 \\ 1+4 & -2-12 \end{pmatrix}$

$AB = \begin{pmatrix} -1 & 5 \\ 5 & -14 \end{pmatrix}$

Step 2: Find the adjoint of $AB$.

For a 2x2 matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the adjoint is $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

Here, $a=-1$, $b=5$, $c=5$, $d=-14$.

$\text{adj}(AB) = \begin{pmatrix} -14 & -5 \\ -5 & -1 \end{pmatrix}$

Right Side: $\text{adj}(B) \text{adj}(A)$

Step 3: Find the adjoint of $A$.

For $A = \begin{pmatrix} 2 & 3 \\ 1 & -4 \end{pmatrix}$, $a=2, b=3, c=1, d=-4$.

$\text{adj}(A) = \begin{pmatrix} -4 & -3 \\ -1 & 2 \end{pmatrix}$

Step 4: Find the adjoint of $B$.

For $B = \begin{pmatrix} 1 & -2 \\ -1 & 3 \end{pmatrix}$, $a=1, b=-2, c=-1, d=3$.

$\text{adj}(B) = \begin{pmatrix} 3 & -(-2) \\ -(-1) & 1 \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix}$

Step 5: Calculate the product $\text{adj}(B) \text{adj}(A)$.

$\text{adj}(B) \text{adj}(A) = \begin{pmatrix} 3 & 2 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} -4 & -3 \\ -1 & 2 \end{pmatrix}$

$= \begin{pmatrix} (3)(-4)+(2)(-1) & (3)(-3)+(2)(2) \\ (1)(-4)+(1)(-1) & (1)(-3)+(1)(2) \end{pmatrix}$

$= \begin{pmatrix} -12-2 & -9+4 \\ -4-1 & -3+2 \end{pmatrix}$

$= \begin{pmatrix} -14 & -5 \\ -5 & -1 \end{pmatrix}$

Step 6: Compare the results.

Left side: $\text{adj}(AB) = \begin{pmatrix} -14 & -5 \\ -5 & -1 \end{pmatrix}$

Right side: $\text{adj}(B) \text{adj}(A) = \begin{pmatrix} -14 & -5 \\ -5 & -1 \end{pmatrix}$

Since the left side equals the right side, the property $\text{adj}(AB) = \text{adj}(B) \text{adj}(A)$ is verified.

Final Answer:

The verification shows that $\text{adj}(AB) = \begin{pmatrix} -14 & -5 \\ -5 & -1 \end{pmatrix}$ and $\text{adj}(B) \text{adj}(A) = \begin{pmatrix} -14 & -5 \\ -5 & -1 \end{pmatrix}$. Thus, $\text{adj}(AB) = \text{adj}(B) \text{adj}(A)$ is verified.

The question asks for the condition under which a square matrix $A$ has an inverse.

Condition for Invertibility:

A square matrix $A$ has an inverse (i.e., $A$ is invertible) if and only if its determinant is non-zero.

Mathematically, this can be stated as:

Other equivalent conditions include:

• The matrix $A$ is non-singular.
• The rows (or columns) of $A$ are linearly independent.
• The rank of $A$ is equal to its order (i.e., $rank(A) = n$ for an $n \times n$ matrix).
• The homogeneous system of equations $AX=0$ has only the trivial solution ($X=0$).
• The adjoint of $A$, $adj(A)$, is non-zero.

However, the most fundamental and commonly used condition is based on the determinant.

Final Answer:

The condition for a square matrix $A$ to have an inverse is that its determinant is non-zero, i.e., $|A| \neq 0$.

The question asks to find the determinant of the adjoint of a matrix $A$, given its order and determinant.

Given:

• $A$ is a $3 \times 3$ matrix.
• $|A| = 5$.

To Find: $|\text{adj}(A)|$.

Solution:

We use the property relating the determinant of the adjoint of a matrix to the determinant of the matrix itself:

where $n$ is the order of the square matrix $A$.

In this problem, we are given:

• $n = 3$ (order of the matrix)
• $|A| = 5$ (determinant of matrix A)

Substitute these values into the formula (i):

$|\text{adj}(A)| = |A|^{3-1}$

$|\text{adj}(A)| = |A|^2$

Now, substitute the value of $|A|$:

$|\text{adj}(A)| = 5^2$

$|\text{adj}(A)| = 25$

Therefore, the determinant of the adjoint of $A$ is 25.

Final Answer:

The value of $|\text{adj}(A)|$ is 25.

The question asks to find the area of a triangle given the coordinates of its vertices.

Given Vertices:

• A = (0, 0)
• B = (5, 0)
• C = (0, 6)

To Find: The area of the triangle ABC.

Solution:

We can find the area of the triangle using the determinant formula for the area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$:

Substitute the given coordinates into the formula:

$Area = \frac{1}{2} \left| \begin{vmatrix} 0 & 0 & 1 \\ 5 & 0 & 1 \\ 0 & 6 & 1 \end{vmatrix} \right|$

To evaluate the determinant, we can expand along the first row, as it contains zeros, which simplifies the calculation:

$D = 0 \cdot C_{11} + 0 \cdot C_{12} + 1 \cdot C_{13}$

$D = 1 \cdot C_{13}$

The cofactor $C_{13}$ is given by $(-1)^{1+3} M_{13}$, where $M_{13}$ is the minor obtained by deleting the first row and third column.

$M_{13} = \begin{vmatrix} 5 & 0 \\ 0 & 6 \end{vmatrix}$

$M_{13} = (5)(6) - (0)(0) = 30 - 0 = 30$

So, $C_{13} = (-1)^{1+3} \times 30 = 1 \times 30 = 30$.

Therefore, the determinant is $D = 1 \times 30 = 30$.

Now, calculate the area:

$Area = \frac{1}{2} |D| = \frac{1}{2} |30| = \frac{1}{2} \times 30 = 15$.

Alternative Geometric Interpretation:

The vertices are A(0, 0) (the origin), B(5, 0) (on the x-axis), and C(0, 6) (on the y-axis).

This forms a right-angled triangle with the right angle at the origin (vertex A).

The base of the triangle can be considered the distance along the x-axis from (0, 0) to (5, 0), which is 5 units.

The height of the triangle can be considered the distance along the y-axis from (0, 0) to (0, 6), which is 6 units.

The area of a triangle is given by $\frac{1}{2} \times \text{base} \times \text{height}$.

$Area = \frac{1}{2} \times 5 \times 6 = \frac{30}{2} = 15$.

Both methods yield the same result.

Final Answer:

The area of the triangle is 15 square units.

The question asks to show that the given determinant is zero using properties of determinants, without direct expansion.

Given Determinant:

$D = \begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix}$

To Prove: $D = 0$ using determinant properties.

Solution:

Let $C_1, C_2, C_3$ represent the columns of the determinant.

Consider the column operation $C_3 \rightarrow C_3 + C_2$. This operation adds the elements of the second column to the corresponding elements of the third column. This operation does not change the value of the determinant.

$D = \begin{vmatrix} 1 & a & (b+c)+a \\ 1 & b & (c+a)+b \\ 1 & c & (a+b)+c \end{vmatrix}$

$D = \begin{vmatrix} 1 & a & a+b+c \\ 1 & b & a+b+c \\ 1 & c & a+b+c \end{vmatrix}$

Now, we can factor out the common term $(a+b+c)$ from the third column:

$D = (a+b+c) \begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix}$

Observe the resulting determinant $\begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix}$. This determinant has two identical columns: the first column ($C_1$) and the third column ($C_3$) are both filled with the value 1.

A property of determinants states that if any two columns (or any two rows) of a determinant are identical, then the value of the determinant is zero.

Since $C_1 = C_3$, the determinant $\begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix} = 0$.

Substituting this back into the expression for $D$:

$D = (a+b+c) \times 0$

$D = 0$

Thus, we have shown that the determinant is equal to 0 using determinant properties.

Final Answer:

The determinant is shown to be 0 by performing the column operation $C_3 \rightarrow C_3 + C_2$, which results in a determinant with two identical columns, making its value zero.

A is an invertible matrix of order $2 \times 2$.

$|A^{-1}| = \frac{1}{|A|}$.

Let A be an invertible matrix of order $2 \times 2$.

By the definition of an invertible matrix, there exists a matrix $A^{-1}$ such that:

Taking the determinant of both sides of the equation, we get:

We know that for any two square matrices A and B of the same order, the determinant of their product is the product of their determinants, i.e., $|AB| = |A||B|$.

Using this property, we have:

The identity matrix of order $2 \times 2$ is $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

The determinant of the identity matrix is:

$|I| = (1)(1) - (0)(0) = 1$

Substituting this value into equation (i), we get:

Since A is an invertible matrix, its determinant is non-zero, i.e., $|A| \neq 0$.

Therefore, we can divide both sides of the equation by $|A|$:

$|A^{-1}| = \frac{1}{|A|}$

Hence, proved.

Let A be a general $2 \times 2$ matrix given by:

$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$

The determinant of A is $|A| = ad - bc$.

Since A is an invertible matrix, we know that its determinant is non-zero, so $|A| = ad - bc \neq 0$.

The inverse of matrix A is given by the formula:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

Where $\text{adj}(A)$ is the adjugate of matrix A. For a $2 \times 2$ matrix, it is given by:

$\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Substituting this into the formula for the inverse, we get:

$A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$

Now, let's find the determinant of $A^{-1}$:

$|A^{-1}| = \left| \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \right|$

Using the property that for a scalar k and a matrix M of order n, $|kM| = k^n |M|$. Here, n=2.

$|A^{-1}| = \left( \frac{1}{ad-bc} \right)^2 \left| \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \right|$

Now, we calculate the determinant of the matrix:

$|A^{-1}| = \frac{1}{(ad-bc)^2} [ (d)(a) - (-b)(-c) ]$

$|A^{-1}| = \frac{1}{(ad-bc)^2} (ad - bc)$

Cancelling one factor of $(ad-bc)$ from the numerator and the denominator:

$|A^{-1}| = \frac{1}{ad-bc}$

Since we know that $|A| = ad-bc$, we can substitute this back into the equation:

$|A^{-1}| = \frac{1}{|A|}$

Hence, proved.

The matrix $A = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}$.

$A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A|I$.

First, we find the determinant of the matrix A, denoted by $|A|$.

$|A| = \left| \begin{matrix} 1 & -2 \\ 3 & 4 \end{matrix} \right|$

$|A| = (1)(4) - (-2)(3)$

$|A| = 4 - (-6)$

$|A| = 4 + 6 = 10$

Next, we find the adjugate of the matrix A, denoted by $\text{adj}(A)$.

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the adjugate is $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

So, for $A = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}$,

$\text{adj}(A) = \begin{pmatrix} 4 & -(-2) \\ -3 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ -3 & 1 \end{pmatrix}$.

Now, let's compute the product $A \cdot \text{adj}(A)$.

$A \cdot \text{adj}(A) = \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 4 & 2 \\ -3 & 1 \end{pmatrix}$

$= \begin{pmatrix} (1)(4)+(-2)(-3) & (1)(2)+(-2)(1) \\ (3)(4)+(4)(-3) & (3)(2)+(4)(1) \end{pmatrix}$

$= \begin{pmatrix} 4+6 & 2-2 \\ 12-12 & 6+4 \end{pmatrix}$

$= \begin{pmatrix} 10 & 0 \\ 0 & 10 \end{pmatrix}$

Next, let's compute the product $\text{adj}(A) \cdot A$.

$\text{adj}(A) \cdot A = \begin{pmatrix} 4 & 2 \\ -3 & 1 \end{pmatrix} \begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix}$

$= \begin{pmatrix} (4)(1)+(2)(3) & (4)(-2)+(2)(4) \\ (-3)(1)+(1)(3) & (-3)(-2)+(1)(4) \end{pmatrix}$

$= \begin{pmatrix} 4+6 & -8+8 \\ -3+3 & 6+4 \end{pmatrix}$

$= \begin{pmatrix} 10 & 0 \\ 0 & 10 \end{pmatrix}$

Finally, let's compute $|A|I$, where I is the identity matrix of order 2.

$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

$|A|I = 10 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$

$= \begin{pmatrix} 10 \times 1 & 10 \times 0 \\ 10 \times 0 & 10 \times 1 \end{pmatrix}$

$= \begin{pmatrix} 10 & 0 \\ 0 & 10 \end{pmatrix}$

From the calculations above, we can see that:

$A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A|I = \begin{pmatrix} 10 & 0 \\ 0 & 10 \end{pmatrix}$

Hence, verified.

Two points on a line are (1, 2) and (3, 6).

The equation of the line passing through the given points using determinants.

Let the two given points be A(1, 2) and B(3, 6).

Let P(x, y) be any point on the line passing through A and B.

If three points are collinear (lie on the same line), the area of the triangle formed by these points is zero.

The area of a triangle with vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ is given by:

Area = $\frac{1}{2} \left| \begin{matrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{matrix} \right|$

For the points P(x, y), A(1, 2), and B(3, 6) to be collinear, the area of $\triangle$PAB must be zero.

So, $\frac{1}{2} \left| \begin{matrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{matrix} \right| = 0$

This implies:

$\left| \begin{matrix} x & y & 1 \\ 1 & 2 & 1 \\ 3 & 6 & 1 \end{matrix} \right| = 0$

Expanding the determinant along the first row (R1):

$x \left| \begin{matrix} 2 & 1 \\ 6 & 1 \end{matrix} \right| - y \left| \begin{matrix} 1 & 1 \\ 3 & 1 \end{matrix} \right| + 1 \left| \begin{matrix} 1 & 2 \\ 3 & 6 \end{matrix} \right| = 0$

Now, we evaluate the $2 \times 2$ determinants:

$x((2)(1) - (6)(1)) - y((1)(1) - (3)(1)) + 1((1)(6) - (3)(2)) = 0$

$x(2 - 6) - y(1 - 3) + 1(6 - 6) = 0$

$x(-4) - y(-2) + 1(0) = 0$

$-4x + 2y = 0$

Dividing the equation by 2, we get:

$-2x + y = 0$

Or, $y = 2x$

This can also be written as $2x - y = 0$.

Therefore, the equation of the line passing through the points (1, 2) and (3, 6) is $y = 2x$.

We can find the equation of the line using the two-point form to verify the result.

The two-point form of a line is given by: $y - y_1 = m(x - x_1)$, where $m$ is the slope.

The slope $m$ is calculated as $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Using the points $(x_1, y_1) = (1, 2)$ and $(x_2, y_2) = (3, 6)$:

$m = \frac{6 - 2}{3 - 1} = \frac{4}{2} = 2$

Now, substituting the slope and one point (1, 2) into the point-slope form:

$y - 2 = 2(x - 1)$

$y - 2 = 2x - 2$

$y = 2x$

This matches the result obtained using the determinant method. Hence, the solution is correct.

A is a skew-symmetric matrix of odd order.

The determinant of A is zero, i.e., $|A| = 0$.

Let A be a skew-symmetric matrix of order $n$, where $n$ is an odd positive integer.

By the definition of a skew-symmetric matrix, we have:

where $A'$ is the transpose of matrix A.

Taking the determinant of both sides of equation (i), we get:

$|A'| = |-A|$

We know two properties of determinants:

1. The determinant of a matrix is equal to the determinant of its transpose, i.e., $|A'| = |A|$.

2. For any $n \times n$ matrix M and any scalar k, $|kM| = k^n |M|$.

Using the second property with $k = -1$, we have:

$|-A| = (-1)^n |A|$

Substituting these properties back into our equation, we get:

It is given that A is a matrix of odd order, which means $n$ is an odd integer.

Therefore, $(-1)^n = -1$.

Substituting this result into equation (ii):

$|A| = -|A|$

Now, we can rearrange the equation:

$|A| + |A| = 0$

$2|A| = 0$

Dividing by 2, we get:

$|A| = 0$

Hence, the determinant of a skew-symmetric matrix of odd order is zero.

Let's verify this property with a general skew-symmetric matrix of order 3 (which is an odd order).

A general skew-symmetric matrix A of order 3 is:

$A = \begin{bmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{bmatrix}$

Note that the diagonal elements are zero, and $a_{ij} = -a_{ji}$.

Now, let's find the determinant of A:

$|A| = \left| \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right|$

Expanding the determinant along the first row (R1):

$|A| = 0 \left| \begin{matrix} 0 & c \\ -c & 0 \end{matrix} \right| - a \left| \begin{matrix} -a & c \\ -b & 0 \end{matrix} \right| + b \left| \begin{matrix} -a & 0 \\ -b & -c \end{matrix} \right|$

$|A| = 0 - a((-a)(0) - (c)(-b)) + b((-a)(-c) - (0)(-b))$

$|A| = -a(0 + bc) + b(ac - 0)$

$|A| = -abc + abc$

$|A| = 0$

This example for a $3 \times 3$ matrix shows that the determinant is zero, thus verifying the general proof.

The matrix $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$.

The adjugate of the transpose of A, which is denoted as $\text{adj}(A')$.

We are given the matrix:

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$

First, we need to find the transpose of the matrix A, denoted as $A'$. The transpose of a matrix is found by interchanging its rows and columns.

$A' = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$

Now, we need to find the adjugate of the matrix $A'$, i.e., $\text{adj}(A')$.

For a general $2 \times 2$ matrix $M = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, its adjugate is given by:

$\text{adj}(M) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$

Applying this formula to our matrix $A' = \begin{pmatrix} 1 & 3 \\ 2 & 4 \end{pmatrix}$, where $a=1, b=3, c=2, d=4$:

$\text{adj}(A') = \begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix}$

Thus, the required matrix is $\text{adj}(A') = \begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix}$.

We can also use the property that for any square matrix A, the adjugate of its transpose is equal to the transpose of its adjugate. That is:

$\text{adj}(A') = (\text{adj}(A))'$

First, we find the adjugate of the original matrix A.

$A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$

Using the formula for the adjugate of a $2 \times 2$ matrix:

$\text{adj}(A) = \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$

Now, we find the transpose of $\text{adj}(A)$.

$(\text{adj}(A))' = \text{transpose of } \begin{pmatrix} 4 & -2 \\ -3 & 1 \end{pmatrix}$

$(\text{adj}(A))' = \begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix}$

Since $\text{adj}(A') = (\text{adj}(A))'$, we get the same result.

Hence, $\text{adj}(A') = \begin{pmatrix} 4 & -3 \\ -2 & 1 \end{pmatrix}$.

$\left| \begin{matrix} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{matrix} \right| = 2(x+y+z)^3$

Let the Left Hand Side (LHS) be $\Delta$.

$\Delta = \left| \begin{matrix} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{matrix} \right|$

Applying the column operation $C_1 \to C_1 + C_2 + C_3$:

$\Delta = \left| \begin{matrix} (x+y+2z)+x+y & x & y \\ z+(y+z+2x)+y & y+z+2x & y \\ z+x+(z+x+2y) & x & z+x+2y \end{matrix} \right|$

$\Delta = \left| \begin{matrix} 2x+2y+2z & x & y \\ 2x+2y+2z & y+z+2x & y \\ 2x+2y+2z & x & z+x+2y \end{matrix} \right|$

Taking $2(x+y+z)$ common from the first column ($C_1$):

$\Delta = 2(x+y+z) \left| \begin{matrix} 1 & x & y \\ 1 & y+z+2x & y \\ 1 & x & z+x+2y \end{matrix} \right|$

Now, to simplify further, let's create zeros in the first column. Applying the row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$:

$\Delta = 2(x+y+z) \left| \begin{matrix} 1 & x & y \\ 1-1 & (y+z+2x)-x & y-y \\ 1-1 & x-x & (z+x+2y)-y \end{matrix} \right|$

$\Delta = 2(x+y+z) \left| \begin{matrix} 1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z \end{matrix} \right|$

Now, expanding the determinant along the first column ($C_1$):

$\Delta = 2(x+y+z) \left[ 1 \left| \begin{matrix} x+y+z & 0 \\ 0 & x+y+z \end{matrix} \right| - 0 \left| \begin{matrix} x & y \\ 0 & x+y+z \end{matrix} \right| + 0 \left| \begin{matrix} x & y \\ x+y+z & 0 \end{matrix} \right| \right]$

$\Delta = 2(x+y+z) [ (x+y+z)(x+y+z) - (0)(0) ]$

$\Delta = 2(x+y+z) [ (x+y+z)^2 ]$

$\Delta = 2(x+y+z)^3$

Thus, LHS = RHS.

Hence, proved.

The matrix $A = \begin{pmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{pmatrix}$.

The inverse of the matrix A, denoted as $A^{-1}$, using the adjoint method.

The formula for the inverse of a matrix A using the adjoint method is:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

First, we need to calculate the determinant of A, $|A|$.

$|A| = \left| \begin{matrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{matrix} \right|$

Expanding along the first row (R1):

$|A| = 1 \left| \begin{matrix} 3 & 0 \\ -2 & 1 \end{matrix} \right| - 2 \left| \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right| + (-2) \left| \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right|$

$|A| = 1((3)(1) - (0)(-2)) - 2((-1)(1) - (0)(0)) - 2((-1)(-2) - (3)(0))$

$|A| = 1(3 - 0) - 2(-1 - 0) - 2(2 - 0)$

$|A| = 1(3) - 2(-1) - 2(2)$

$|A| = 3 + 2 - 4 = 1$

Since $|A| = 1 \neq 0$, the matrix is invertible and its inverse exists.

Next, we find the adjugate of A, which is the transpose of the cofactor matrix.

The cofactors are calculated as follows:

$C_{11} = (-1)^{1+1} \left| \begin{matrix} 3 & 0 \\ -2 & 1 \end{matrix} \right| = 1(3-0) = 3$

$C_{12} = (-1)^{1+2} \left| \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right| = -1(-1-0) = 1$

$C_{13} = (-1)^{1+3} \left| \begin{matrix} -1 & 3 \\ 0 & -2 \end{matrix} \right| = 1(2-0) = 2$

$C_{21} = (-1)^{2+1} \left| \begin{matrix} 2 & -2 \\ -2 & 1 \end{matrix} \right| = -1(2-4) = 2$

$C_{22} = (-1)^{2+2} \left| \begin{matrix} 1 & -2 \\ 0 & 1 \end{matrix} \right| = 1(1-0) = 1$

$C_{23} = (-1)^{2+3} \left| \begin{matrix} 1 & 2 \\ 0 & -2 \end{matrix} \right| = -1(-2-0) = 2$

$C_{31} = (-1)^{3+1} \left| \begin{matrix} 2 & -2 \\ 3 & 0 \end{matrix} \right| = 1(0-(-6)) = 6$

$C_{32} = (-1)^{3+2} \left| \begin{matrix} 1 & -2 \\ -1 & 0 \end{matrix} \right| = -1(0-2) = 2$

$C_{33} = (-1)^{3+3} \left| \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right| = 1(3-(-2)) = 5$

The cofactor matrix is:

Cofactor(A) = $\begin{pmatrix} 3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5 \end{pmatrix}$

The adjugate of A is the transpose of the cofactor matrix:

$\text{adj}(A) = (\text{Cofactor(A)})' = \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix}$

Now, we can find the inverse using the formula:

$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1} \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix}$

$A^{-1} = \begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix}$

Therefore, the inverse of the matrix A is $\begin{pmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{pmatrix}$.

The system of linear equations is:

$x - y + 2z = 7$

$3x + 4y - 5z = -5$

$2x - y + 3z = 12$

The solution for x, y, and z using the matrix method.

The given system of linear equations can be written in the matrix form $AX = B$, where:

$A = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, and $B = \begin{pmatrix} 7 \\ -5 \\ 12 \end{pmatrix}$

The solution is given by $X = A^{-1}B$. First, we need to find the inverse of matrix A.

The inverse is given by $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Step 1: Calculate the determinant of A, $|A|$.

$|A| = \left| \begin{matrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{matrix} \right|$

$|A| = 1(4 \cdot 3 - (-5) \cdot (-1)) - (-1)(3 \cdot 3 - (-5) \cdot 2) + 2(3 \cdot (-1) - 4 \cdot 2)$

$|A| = 1(12 - 5) + 1(9 + 10) + 2(-3 - 8)$

$|A| = 1(7) + 1(19) + 2(-11)$

$|A| = 7 + 19 - 22 = 4$

Since $|A| = 4 \neq 0$, a unique solution exists.

Step 2: Find the adjugate of A, adj(A).

First, we find the cofactors of each element:

$C_{11} = (12-5) = 7$

$C_{12} = -(9 - (-10)) = -19$

$C_{13} = (-3 - 8) = -11$

$C_{21} = -(-3 - (-2)) = 1$

$C_{22} = (3 - 4) = -1$

$C_{23} = -(-1 - (-2)) = -1$

$C_{31} = (5 - 8) = -3$

$C_{32} = -(-5 - 6) = 11$

$C_{33} = (4 - (-3)) = 7$

The cofactor matrix is $\begin{pmatrix} 7 & -19 & -11 \\ 1 & -1 & -1 \\ -3 & 11 & 7 \end{pmatrix}$.

The adjugate of A is the transpose of the cofactor matrix:

$\text{adj}(A) = \begin{pmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{pmatrix}$

Step 3: Calculate the inverse of A, $A^{-1}$.

$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{4} \begin{pmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{pmatrix}$

Step 4: Solve for X.

$X = A^{-1}B = \frac{1}{4} \begin{pmatrix} 7 & 1 & -3 \\ -19 & -1 & 11 \\ -11 & -1 & 7 \end{pmatrix} \begin{pmatrix} 7 \\ -5 \\ 12 \end{pmatrix}$

$X = \frac{1}{4} \begin{pmatrix} 7(7) + 1(-5) + (-3)(12) \\ -19(7) + (-1)(-5) + 11(12) \\ -11(7) + (-1)(-5) + 7(12) \end{pmatrix}$

$X = \frac{1}{4} \begin{pmatrix} 49 - 5 - 36 \\ -133 + 5 + 132 \\ -77 + 5 + 84 \end{pmatrix}$

$X = \frac{1}{4} \begin{pmatrix} 8 \\ 4 \\ 12 \end{pmatrix}$

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 8/4 \\ 4/4 \\ 12/4 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$

By comparing the elements, we get the solution:

$x = 2$, $y = 1$, and $z = 3$.

$\left| \begin{matrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{matrix} \right| = abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$

Let the Left Hand Side (LHS) be $\Delta$.

$\Delta = \left| \begin{matrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{matrix} \right|$

Taking $a$, $b$, and $c$ as common factors from rows $R_1$, $R_2$, and $R_3$ respectively (assuming $a,b,c \neq 0$):

$\Delta = abc \left| \begin{matrix} \frac{1+a}{a} & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1+b}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1+c}{c} \end{matrix} \right|$

$\Delta = abc \left| \begin{matrix} 1+\frac{1}{a} & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & 1+\frac{1}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & 1+\frac{1}{c} \end{matrix} \right|$

Applying the row operation $R_1 \to R_1 + R_2 + R_3$:

$\Delta = abc \left| \begin{matrix} 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\ \frac{1}{b} & 1+\frac{1}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & 1+\frac{1}{c} \end{matrix} \right|$

Taking the common factor $\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$ from the first row:

$\Delta = abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \left| \begin{matrix} 1 & 1 & 1 \\ \frac{1}{b} & 1+\frac{1}{b} & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & 1+\frac{1}{c} \end{matrix} \right|$

Now, to simplify the determinant, we apply column operations $C_2 \to C_2 - C_1$ and $C_3 \to C_3 - C_1$:

$\Delta = abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \left| \begin{matrix} 1 & 1-1 & 1-1 \\ \frac{1}{b} & (1+\frac{1}{b})-\frac{1}{b} & \frac{1}{b}-\frac{1}{b} \\ \frac{1}{c} & \frac{1}{c}-\frac{1}{c} & (1+\frac{1}{c})-\frac{1}{c} \end{matrix} \right|$

$\Delta = abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \left| \begin{matrix} 1 & 0 & 0 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1 \end{matrix} \right|$

Expanding the determinant along the first row:

$\Delta = abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) \left[ 1 \left| \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right| - 0 + 0 \right]$

$\Delta = abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) [1(1-0)]

$\Delta = abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$

Thus, LHS = RHS.

Hence, proved.

Let the LHS be $\Delta$.

$\Delta = \left| \begin{matrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{matrix} \right|$

Applying row operations $R_1 \to R_1 - R_2$ and $R_2 \to R_2 - R_3$:

$\Delta = \left| \begin{matrix} (1+a)-1 & 1-(1+b) & 1-1 \\ 1-1 & (1+b)-1 & 1-(1+c) \\ 1 & 1 & 1+c \end{matrix} \right|$

$\Delta = \left| \begin{matrix} a & -b & 0 \\ 0 & b & -c \\ 1 & 1 & 1+c \end{matrix} \right|$

Now, expanding the determinant along the first row:

$\Delta = a \left| \begin{matrix} b & -c \\ 1 & 1+c \end{matrix} \right| - (-b) \left| \begin{matrix} 0 & -c \\ 1 & 1+c \end{matrix} \right| + 0 \left| \begin{matrix} 0 & b \\ 1 & 1 \end{matrix} \right|$

$\Delta = a[b(1+c) - (-c)(1)] + b[0(1+c) - (-c)(1)] + 0$

$\Delta = a(b+bc+c) + b(c)$

$\Delta = ab + abc + ac + bc$

Now, let's simplify the Right Hand Side (RHS):

RHS $= abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$

RHS $= abc \cdot 1 + abc \cdot \frac{1}{a} + abc \cdot \frac{1}{b} + abc \cdot \frac{1}{c}$

RHS $= abc + bc + ac + ab$

Since LHS = $ab + bc + ca + abc$ and RHS = $ab + bc + ca + abc$, we have LHS = RHS.

Hence, proved.

The matrix $A = \begin{pmatrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{pmatrix}$.

The inverse of the matrix A, denoted as $A^{-1}$, using the adjoint method.

The formula for the inverse of a matrix A using the adjoint method is given by:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

First, we need to calculate the determinant of A, denoted as $|A|$.

$|A| = \left| \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right|$

Expanding the determinant along the first row (R1):

$|A| = 2 \left| \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right| - (-1) \left| \begin{matrix} -1 & -1 \\ 1 & 2 \end{matrix} \right| + 1 \left| \begin{matrix} -1 & 2 \\ 1 & -1 \end{matrix} \right|$

$|A| = 2((2)(2) - (-1)(-1)) + 1((-1)(2) - (-1)(1)) + 1((-1)(-1) - (2)(1))$

$|A| = 2(4 - 1) + 1(-2 + 1) + 1(1 - 2)$

$|A| = 2(3) + 1(-1) + 1(-1)$

$|A| = 6 - 1 - 1 = 4$

Since $|A| = 4 \neq 0$, the matrix is invertible.

Next, we find the adjugate of A, which is the transpose of the cofactor matrix.

The cofactors of the elements of A are:

$C_{11} = (-1)^{1+1} \left| \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right| = 1(4 - 1) = 3$

$C_{12} = (-1)^{1+2} \left| \begin{matrix} -1 & -1 \\ 1 & 2 \end{matrix} \right| = -1(-2 - (-1)) = -1(-1) = 1$

$C_{13} = (-1)^{1+3} \left| \begin{matrix} -1 & 2 \\ 1 & -1 \end{matrix} \right| = 1(1 - 2) = -1$

$C_{21} = (-1)^{2+1} \left| \begin{matrix} -1 & 1 \\ -1 & 2 \end{matrix} \right| = -1(-2 - (-1)) = -1(-1) = 1$

$C_{22} = (-1)^{2+2} \left| \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} \right| = 1(4 - 1) = 3$

$C_{23} = (-1)^{2+3} \left| \begin{matrix} 2 & -1 \\ 1 & -1 \end{matrix} \right| = -1(-2 - (-1)) = -1(-1) = 1$

$C_{31} = (-1)^{3+1} \left| \begin{matrix} -1 & 1 \\ 2 & -1 \end{matrix} \right| = 1(1 - 2) = -1$

$C_{32} = (-1)^{3+2} \left| \begin{matrix} 2 & 1 \\ -1 & -1 \end{matrix} \right| = -1(-2 - (-1)) = -1(-1) = 1$

$C_{33} = (-1)^{3+3} \left| \begin{matrix} 2 & -1 \\ -1 & 2 \end{matrix} \right| = 1(4 - 1) = 3$

The cofactor matrix of A is:

Cofactor(A) = $\begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}$

The adjugate of A is the transpose of the cofactor matrix:

$\text{adj}(A) = (\text{Cofactor(A)})' = \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}$

Now, we can find the inverse of A:

$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{4} \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}$

Therefore, the inverse of the matrix A is $\frac{1}{4} \begin{pmatrix} 3 & 1 & -1 \\ 1 & 3 & 1 \\ -1 & 1 & 3 \end{pmatrix}$.

The system of linear equations:

$2x + 3y = 5$

$x - 2y = 1$

The solution for x and y using Cramer's rule.

The given system of linear equations is:

According to Cramer's rule, the solution to a system of equations $a_1x + b_1y = d_1$ and $a_2x + b_2y = d_2$ is given by:

$x = \frac{D_x}{D}$ and $y = \frac{D_y}{D}$, provided $D \neq 0$.

First, we find the determinant of the coefficient matrix, D.

$D = \left| \begin{matrix} 2 & 3 \\ 1 & -2 \end{matrix} \right|$

$D = (2)(-2) - (3)(1)$

$D = -4 - 3 = -7$

Since $D = -7 \neq 0$, a unique solution exists.

Next, we find the determinant $D_x$ by replacing the column of x-coefficients in D with the constant terms.

$D_x = \left| \begin{matrix} 5 & 3 \\ 1 & -2 \end{matrix} \right|$

$D_x = (5)(-2) - (3)(1)$

$D_x = -10 - 3 = -13$

Next, we find the determinant $D_y$ by replacing the column of y-coefficients in D with the constant terms.

$D_y = \left| \begin{matrix} 2 & 5 \\ 1 & 1 \end{matrix} \right|$

$D_y = (2)(1) - (5)(1)$

$D_y = 2 - 5 = -3$

Now, we can calculate the values of x and y.

$x = \frac{D_x}{D} = \frac{-13}{-7} = \frac{13}{7}$

$y = \frac{D_y}{D} = \frac{-3}{-7} = \frac{3}{7}$

Therefore, the solution to the system of equations is $x = \frac{13}{7}$ and $y = \frac{3}{7}$.

We can verify the solution by substituting the values of x and y back into one of the original equations.

Using the second equation, $x - 2y = 1$:

LHS $= x - 2y = \frac{13}{7} - 2\left(\frac{3}{7}\right)$

$= \frac{13}{7} - \frac{6}{7}$

$= \frac{13-6}{7}$

$= \frac{7}{7} = 1$

RHS $= 1$

Since LHS = RHS, our solution is correct.

$\left| \begin{matrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{matrix} \right| = 4abc$

Let the Left Hand Side (LHS) be $\Delta$.

$\Delta = \left| \begin{matrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{matrix} \right|$

Applying the row operation $R_1 \to R_1 - R_2 - R_3$:

The first element of the new first row is $(b+c) - b - c = 0$.

The second element of the new first row is $a - (c+a) - c = a - c - a - c = -2c$.

The third element of the new first row is $a - b - (a+b) = a - b - a - b = -2b$.

So, the determinant becomes:

$\Delta = \left| \begin{matrix} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \end{matrix} \right|$

Now, expanding the determinant along the first row ($R_1$):

$\Delta = 0 \left| \begin{matrix} c+a & b \\ c & a+b \end{matrix} \right| - (-2c) \left| \begin{matrix} b & b \\ c & a+b \end{matrix} \right| + (-2b) \left| \begin{matrix} b & c+a \\ c & c \end{matrix} \right|$

$\Delta = 0 + 2c [b(a+b) - b(c)] - 2b [b(c) - c(c+a)]$

$\Delta = 2c(ab + b^2 - bc) - 2b(bc - c^2 - ac)$

$\Delta = 2abc + 2b^2c - 2bc^2 - 2b^2c + 2bc^2 + 2abc$

The terms $2b^2c$ and $-2b^2c$ cancel out. The terms $-2bc^2$ and $2bc^2$ cancel out.

$\Delta = 2abc + 2abc$

$\Delta = 4abc$

Thus, LHS = RHS.

Hence, proved.

Let $\Delta = \left| \begin{matrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{matrix} \right|$.

The determinant is a polynomial in $a, b, c$. Let's check the value of the determinant when $a=0$.

If $a=0$, $\Delta = \left| \begin{matrix} b+c & 0 & 0 \\ b & c & b \\ c & c & b \end{matrix} \right|$.

Expanding along R1: $\Delta = (b+c) \left| \begin{matrix} c & b \\ c & b \end{matrix} \right| = (b+c)(cb - cb) = 0$.

Since $\Delta = 0$ when $a=0$, $(a-0)$ or $a$ is a factor of $\Delta$.

By symmetry, if we set $b=0$ or $c=0$, the determinant will also become zero. Thus, $b$ and $c$ are also factors of $\Delta$.

Therefore, $abc$ is a factor of $\Delta$.

Since the determinant is a homogeneous polynomial of degree 3, we can write $\Delta = k \cdot abc$ for some constant $k$.

To find the value of $k$, we can substitute any non-zero values for $a, b, c$. Let's take $a=1, b=1, c=1$.

LHS = $\Delta = \left| \begin{matrix} 1+1 & 1 & 1 \\ 1 & 1+1 & 1 \\ 1 & 1 & 1+1 \end{matrix} \right| = \left| \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{matrix} \right|$

$= 2(2 \cdot 2 - 1 \cdot 1) - 1(1 \cdot 2 - 1 \cdot 1) + 1(1 \cdot 1 - 2 \cdot 1)$

$= 2(4 - 1) - 1(2 - 1) + 1(1 - 2)$

$= 2(3) - 1(1) + 1(-1)$

$= 6 - 1 - 1 = 4$

RHS = $k \cdot abc = k \cdot (1)(1)(1) = k$.

Equating LHS and RHS, we get $k=4$.

Therefore, $\Delta = 4abc$.

Hence, proved.

The matrix $A = \begin{pmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{pmatrix}$.

The system of linear equations:

$x + 2y = 1$

$-2x - y - 2z = -2$

$-y + z = 0$

The inverse of matrix A ($A^{-1}$) and the solution for x, y, and z.

Part 1: Finding the inverse of A ($A^{-1}$)

The inverse is given by the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Step 1: Calculate the determinant of A, $|A|$.

$|A| = \left| \begin{matrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{matrix} \right|$

$|A| = 1((-1)(1) - (-2)(-1)) - 2((-2)(1) - (-2)(0)) + 0$

$|A| = 1(-1 - 2) - 2(-2 - 0)$

$|A| = 1(-3) - 2(-2) = -3 + 4 = 1$

Since $|A| = 1 \neq 0$, the inverse exists.

Step 2: Find the adjugate of A, adj(A).

We find the cofactors of each element:

$C_{11} = (-1-2) = -3$

$C_{12} = -(-2-0) = 2$

$C_{13} = (2-0) = 2$

$C_{21} = -(2-0) = -2$

$C_{22} = (1-0) = 1$

$C_{23} = -(-1-0) = 1$

$C_{31} = (-4-0) = -4$

$C_{32} = -(-2-0) = 2$

$C_{33} = (-1-(-4)) = 3$

The cofactor matrix is $\begin{pmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{pmatrix}$.

The adjugate of A is the transpose of the cofactor matrix:

$\text{adj}(A) = \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{pmatrix}$

Step 3: Calculate the inverse of A, $A^{-1}$.

$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{1} \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{pmatrix}$

$A^{-1} = \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{pmatrix}$

Part 2: Solving the system of equations

The given system of linear equations is:

$x + 2y + 0z = 1$

$-2x - y - 2z = -2$

$0x - y + z = 0$

This system can be written in the matrix form $A'X = B$, where:

$A' = \begin{pmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, and $B = \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$

We notice that the coefficient matrix $A'$ is the same as the given matrix A.

So, the system is $AX = B$.

The solution is given by $X = A^{-1}B$. We have already calculated $A^{-1}$.

$X = \begin{pmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{pmatrix} \begin{pmatrix} 1 \\ -2 \\ 0 \end{pmatrix}$

$X = \begin{pmatrix} (-3)(1) + (-2)(-2) + (-4)(0) \\ (2)(1) + (1)(-2) + (2)(0) \\ (2)(1) + (1)(-2) + (3)(0) \end{pmatrix}$

$X = \begin{pmatrix} -3 + 4 + 0 \\ 2 - 2 + 0 \\ 2 - 2 + 0 \end{pmatrix}$

$X = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$

By comparing the elements, we get the solution:

$x = 1$, $y = 0$, and $z = 0$.

The system of linear equations is:

$x + y + z = 6$

$x + 2y + 3z = 14$

$x + 4y + 7z = 30$

The solution for the system of equations using the matrix method.

The given system of linear equations can be written in the matrix form $AX = B$, where:

$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, and $B = \begin{pmatrix} 6 \\ 14 \\ 30 \end{pmatrix}$

First, we calculate the determinant of the coefficient matrix A.

$|A| = \left| \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 4 & 7 \end{matrix} \right|$

$|A| = 1(2 \cdot 7 - 3 \cdot 4) - 1(1 \cdot 7 - 3 \cdot 1) + 1(1 \cdot 4 - 2 \cdot 1)$

$|A| = 1(14 - 12) - 1(7 - 3) + 1(4 - 2)$

$|A| = 1(2) - 1(4) + 1(2)$

$|A| = 2 - 4 + 2 = 0$

Since the determinant of A is zero, $|A| = 0$, the system of equations either has no solution (inconsistent) or has infinitely many solutions.

To determine which case it is, we need to find $(\text{adj } A)B$.

First, we find the adjugate of A, $\text{adj}(A)$. The cofactors are:

$C_{11} = (14-12) = 2$

$C_{12} = -(7-3) = -4$

$C_{13} = (4-2) = 2$

$C_{21} = -(7-4) = -3$

$C_{22} = (7-1) = 6$

$C_{23} = -(4-1) = -3$

$C_{31} = (3-2) = 1$

$C_{32} = -(3-1) = -2$

$C_{33} = (2-1) = 1$

The cofactor matrix is $\begin{pmatrix} 2 & -4 & 2 \\ -3 & 6 & -3 \\ 1 & -2 & 1 \end{pmatrix}$.

The adjugate of A is the transpose of the cofactor matrix:

$\text{adj}(A) = \begin{pmatrix} 2 & -3 & 1 \\ -4 & 6 & -2 \\ 2 & -3 & 1 \end{pmatrix}$

Now, we compute the product $(\text{adj } A)B$:

$(\text{adj } A)B = \begin{pmatrix} 2 & -3 & 1 \\ -4 & 6 & -2 \\ 2 & -3 & 1 \end{pmatrix} \begin{pmatrix} 6 \\ 14 \\ 30 \end{pmatrix}$

$(\text{adj } A)B = \begin{pmatrix} 2(6) - 3(14) + 1(30) \\ -4(6) + 6(14) - 2(30) \\ 2(6) - 3(14) + 1(30) \end{pmatrix}$

$(\text{adj } A)B = \begin{pmatrix} 12 - 42 + 30 \\ -24 + 84 - 60 \\ 12 - 42 + 30 \end{pmatrix}$

$(\text{adj } A)B = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} = O$ (the zero matrix)

Since $|A| = 0$ and $(\text{adj } A)B = O$, the system of equations is consistent and has infinitely many solutions.

To find the general solution, we let $z = k$, where k is any real number.

Substitute $z = k$ into the first two equations:

Subtracting equation (i) from equation (ii):

$(x+2y) - (x+y) = (14-3k) - (6-k)$

$y = 14 - 3k - 6 + k$

$y = 8 - 2k$

Substitute the value of y into equation (i):

$x + (8 - 2k) = 6 - k$

$x = 6 - k - 8 + 2k$

$x = k - 2$

Thus, the system has infinitely many solutions given by:

$x = k - 2$

$y = 8 - 2k$

$z = k$

where k is any real number.

$\left| \begin{matrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{matrix} \right| = (x-y)(y-z)(z-x)$

Let the Left Hand Side (LHS) be denoted by $\Delta$.

$\Delta = \left| \begin{matrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{matrix} \right|$

To simplify the determinant, we aim to create zeros. We can do this by applying row operations.

$\Delta = \left| \begin{matrix} 1-1 & x-y & x^2-y^2 \\ 1-1 & y-z & y^2-z^2 \\ 1 & z & z^2 \end{matrix} \right|$

$\Delta = \left| \begin{matrix} 0 & x-y & (x-y)(x+y) \\ 0 & y-z & (y-z)(y+z) \\ 1 & z & z^2 \end{matrix} \right|$

Now, we can take the common factor $(x-y)$ from the first row ($R_1$) and $(y-z)$ from the second row ($R_2$).

$\Delta = (x-y)(y-z) \left| \begin{matrix} 0 & 1 & x+y \\ 0 & 1 & y+z \\ 1 & z & z^2 \end{matrix} \right|$

Now, we expand the determinant along the first column ($C_1$), as it contains two zeros, which simplifies the calculation.

$\Delta = (x-y)(y-z) \left[ 0 \left| \begin{matrix} 1 & y+z \\ z & z^2 \end{matrix} \right| - 0 \left| \begin{matrix} 1 & x+y \\ z & z^2 \end{matrix} \right| + 1 \left| \begin{matrix} 1 & x+y \\ 1 & y+z \end{matrix} \right| \right]$

$\Delta = (x-y)(y-z) [ 0 - 0 + 1( (1)(y+z) - (1)(x+y) ) ]$

$\Delta = (x-y)(y-z) (y+z - x-y)$

$\Delta = (x-y)(y-z) (z-x)$

This is equal to the Right Hand Side (RHS).

Hence, proved.

The matrix $A = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}$.

The system of linear equations:

$2x - 3y + 5z = 16$

$3x + 2y - 4z = -4$

$x + y - 2z = -3$

The inverse of matrix A ($A^{-1}$) and the solution for x, y, and z.

Part 1: Finding the inverse of A ($A^{-1}$)

The inverse is given by the formula $A^{-1} = \frac{1}{|A|} \text{adj}(A)$.

Step 1: Calculate the determinant of A, $|A|$.

$|A| = \left| \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right|$

$|A| = 2((2)(-2) - (-4)(1)) - (-3)((3)(-2) - (-4)(1)) + 5((3)(1) - (2)(1))$

$|A| = 2(-4 + 4) + 3(-6 + 4) + 5(3 - 2)$

$|A| = 2(0) + 3(-2) + 5(1)$

$|A| = 0 - 6 + 5 = -1$

Since $|A| = -1 \neq 0$, the inverse exists.

Step 2: Find the adjugate of A, adj(A).

We find the cofactors of each element:

$C_{11} = (2)(-2) - (-4)(1) = -4 + 4 = 0$

$C_{12} = -((3)(-2) - (-4)(1)) = -(-6 + 4) = 2$

$C_{13} = (3)(1) - (2)(1) = 3 - 2 = 1$

$C_{21} = -((-3)(-2) - (5)(1)) = -(6 - 5) = -1$

$C_{22} = (2)(-2) - (5)(1) = -4 - 5 = -9$

$C_{23} = -((2)(1) - (-3)(1)) = -(2 + 3) = -5$

$C_{31} = (-3)(-4) - (5)(2) = 12 - 10 = 2$

$C_{32} = -((2)(-4) - (5)(3)) = -(-8 - 15) = 23$

$C_{33} = (2)(2) - (-3)(3) = 4 + 9 = 13$

The cofactor matrix is $\begin{pmatrix} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{pmatrix}$.

The adjugate of A is the transpose of the cofactor matrix:

$\text{adj}(A) = \begin{pmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{pmatrix}$

Step 3: Calculate the inverse of A, $A^{-1}$.

$A^{-1} = \frac{1}{|A|} \text{adj}(A) = \frac{1}{-1} \begin{pmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{pmatrix}$

$A^{-1} = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix}$

Part 2: Solving the system of equations

The given system of linear equations is:

$2x - 3y + 5z = 16$

$3x + 2y - 4z = -4$

$x + y - 2z = -3$

This system can be written in the matrix form $A'X = B$, where:

$A' = \begin{pmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{pmatrix}$, $X = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$, and $B = \begin{pmatrix} 16 \\ -4 \\ -3 \end{pmatrix}$

We notice that the coefficient matrix $A'$ is the same as the given matrix A.

So, the system is $AX = B$.

The solution is given by $X = A^{-1}B$. We have already calculated $A^{-1}$.

$X = \begin{pmatrix} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{pmatrix} \begin{pmatrix} 16 \\ -4 \\ -3 \end{pmatrix}$

$X = \begin{pmatrix} (0)(16) + (1)(-4) + (-2)(-3) \\ (-2)(16) + (9)(-4) + (-23)(-3) \\ (-1)(16) + (5)(-4) + (-13)(-3) \end{pmatrix}$

$X = \begin{pmatrix} 0 - 4 + 6 \\ -32 - 36 + 69 \\ -16 - 20 + 39 \end{pmatrix}$

$X = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$

$\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 3 \end{pmatrix}$

By comparing the elements, we get the solution:

$x = 2$, $y = 1$, and $z = 3$.

The matrix $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$.

1. Verify that $A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A|I$.

2. Hence find $A^{-1}$.

First, we calculate the determinant of A, denoted as $|A|$.

$|A| = \left| \begin{matrix} 2 & -1 \\ 3 & 4 \end{matrix} \right| = (2)(4) - (-1)(3) = 8 + 3 = 11$.

Next, we find the adjugate of A, denoted as $\text{adj}(A)$.

For a $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$, the adjugate is $\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}$.

So, for $A = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$, we have:

$\text{adj}(A) = \begin{pmatrix} 4 & -(-1) \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix}$.

Now, we compute the product $A \cdot \text{adj}(A)$:

$A \cdot \text{adj}(A) = \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix} \begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix}$

$= \begin{pmatrix} (2)(4)+(-1)(-3) & (2)(1)+(-1)(2) \\ (3)(4)+(4)(-3) & (3)(1)+(4)(2) \end{pmatrix}$

$= \begin{pmatrix} 8+3 & 2-2 \\ 12-12 & 3+8 \end{pmatrix} = \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}$.

Next, we compute the product $\text{adj}(A) \cdot A$:

$\text{adj}(A) \cdot A = \begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix} \begin{pmatrix} 2 & -1 \\ 3 & 4 \end{pmatrix}$

$= \begin{pmatrix} (4)(2)+(1)(3) & (4)(-1)+(1)(4) \\ (-3)(2)+(2)(3) & (-3)(-1)+(2)(4) \end{pmatrix}$

$= \begin{pmatrix} 8+3 & -4+4 \\ -6+6 & 3+8 \end{pmatrix} = \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}$.

Finally, we compute $|A|I$, where I is the identity matrix of order 2.

$|A|I = 11 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 11 \times 1 & 11 \times 0 \\ 11 \times 0 & 11 \times 1 \end{pmatrix} = \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}$.

From the calculations, we can see that:

$A \cdot \text{adj}(A) = \text{adj}(A) \cdot A = |A|I = \begin{pmatrix} 11 & 0 \\ 0 & 11 \end{pmatrix}$.

Hence, verified.

From the verified relationship, we have:

Since $|A| = 11 \neq 0$, A is invertible.

We can pre-multiply both sides of equation (i) by $A^{-1}$:

$A^{-1}(A \cdot \text{adj}(A)) = A^{-1}(|A|I)$

Using associativity of matrix multiplication, $(A^{-1}A) \cdot \text{adj}(A) = |A|(A^{-1}I)$.

Since $A^{-1}A = I$ and $A^{-1}I = A^{-1}$, the equation becomes:

$I \cdot \text{adj}(A) = |A|A^{-1}$

$\text{adj}(A) = |A|A^{-1}$

Dividing by the scalar $|A|$, we get the formula for the inverse:

$A^{-1} = \frac{1}{|A|} \text{adj}(A)$

Substituting the values we found earlier:

$A^{-1} = \frac{1}{11} \begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix}$

Therefore, the inverse of the matrix A is $A^{-1} = \frac{1}{11} \begin{pmatrix} 4 & 1 \\ -3 & 2 \end{pmatrix}$.