Matrices: Properties and Advanced Operations

Transpose of a Matrix

The transpose of a matrix is one of the fundamental operations in linear algebra. It involves creating a new matrix by simply swapping the rows and columns of the original matrix. This means that the first row of the original matrix becomes the first column of the transposed matrix, the second row becomes the second column, and so on.

More formally, if $A$ is a matrix of order $m \times n$ (meaning it has $m$ rows and $n$ columns), its transpose is denoted by $A^T$ or sometimes $A'$. The process of taking the transpose results in a matrix with the number of rows and columns interchanged. Thus, the order of the transposed matrix $A^T$ will be $n \times m$ (it will have $n$ rows and $m$ columns).

If the element located at the $i$-th row and $j$-th column of the original matrix $A$ is denoted by $a_{ij}$, then the element located at the $j$-th row and $i$-th column of the transposed matrix $A^T$ (denoted as $a'_{ji}$) is equal to the element $a_{ij}$ from the original matrix. Symbolically:

Note that the element in the $j$-th row and $i$-th column of $A^T$ is the element from the $i$-th row and $j$-th column of $A$.

For a square matrix $A$ of order $n \times n$, its transpose $A^T$ will also be of order $n \times n$.

Example of Transpose

Answer:

Properties of Transpose of a Matrix

The transpose operation interacts with other matrix operations (addition, multiplication, scalar multiplication) in specific ways. Let $A$ and $B$ be matrices, and $k$ be a scalar. Assume the dimensions of $A$ and $B$ are such that the indicated operations are defined (e.g., same order for addition/subtraction, compatible orders for multiplication).

Property 1: Transpose of the Transpose

Taking the transpose of a matrix twice returns the original matrix.

Derivation: Let $A = [a_{ij}]$ be an $m \times n$ matrix. Its transpose is $A^T = [a'_{ji}]$, where $a'_{ji} = a_{ij}$ for $1 \le i \le m, 1 \le j \le n$. $A^T$ is an $n \times m$ matrix. Now, take the transpose of $A^T$. Let $(A^T)^T = [a''_{ik}]$, where $a''_{ik}$ is the element in the $i$-th row and $k$-th column of $(A^T)^T$. By the definition of transpose applied to $A^T$, the element $a''_{ik}$ is the element in the $k$-th row and $i$-th column of $A^T$. So, $a''_{ik} = a'_{ki}$. By the definition of $A^T$, $a'_{ki} = a_{ik}$. Thus, $a''_{ik} = a_{ik}$. Since the element in the $i$-th row and $k$-th column of $(A^T)^T$ is equal to the element in the $i$-th row and $k$-th column of $A$ for all $1 \le i \le m, 1 \le k \le n$, and both matrices have the same order $m \times n$, it follows that $(A^T)^T = A$.

Property 2: Transpose of a Scalar Multiple

The transpose of a scalar multiple of a matrix is equal to the scalar multiple of the transpose of the matrix.

Derivation: Let $A = [a_{ij}]$ be an $m \times n$ matrix and $k$ be a scalar. The matrix $kA = [ka_{ij}]$ is an $m \times n$ matrix. The element in the $i$-th row and $j$-th column of $kA$ is $ka_{ij}$. The transpose $(kA)^T = [b_{ji}]$ is an $n \times m$ matrix, where $b_{ji}$ is the element in the $j$-th row and $i$-th column of $(kA)^T$. By definition of transpose, $b_{ji} = ka_{ij}$. Now consider $kA^T$. The transpose $A^T = [a'_{ji}]$ is an $n \times m$ matrix, where $a'_{ji} = a_{ij}$. The scalar multiple $kA^T = [k a'_{ji}]$ is also an $n \times m$ matrix. The element in the $j$-th row and $i$-th column of $kA^T$ is $k a'_{ji} = k a_{ij}$. Since the elements in the $j$-th row and $i$-th column of $(kA)^T$ and $kA^T$ are both equal to $ka_{ij}$ for all $j, i$, and both matrices have the same order ($n \times m$), we have $(kA)^T = kA^T$.

Property 3: Transpose of a Sum or Difference

The transpose of the sum or difference of two matrices is the sum or difference of their transposes. This property holds if the matrices $A$ and $B$ have the same order so that their sum or difference is defined.

Derivation: Let $A = [a_{ij}]$ and $B = [b_{ij}]$ be two $m \times n$ matrices. Their sum (or difference) is $A \pm B = [a_{ij} \pm b_{ij}]$, which is an $m \times n$ matrix. The element in the $i$-th row and $j$-th column of $(A \pm B)$ is $a_{ij} \pm b_{ij}$. The transpose $(A \pm B)^T = [c_{ji}]$ is an $n \times m$ matrix, where $c_{ji}$ is the element in the $j$-th row and $i$-th column of $(A \pm B)^T$. By definition of transpose, $c_{ji} = a_{ij} \pm b_{ij}$. Now consider $A^T \pm B^T$. $A^T = [a'_{ji}]$ is an $n \times m$ matrix, where $a'_{ji} = a_{ij}$. $B^T = [b'_{ji}]$ is an $n \times m$ matrix, where $b'_{ji} = b_{ij}$. The sum (or difference) $A^T \pm B^T = [a'_{ji} \pm b'_{ji}]$ is an $n \times m$ matrix. The element in the $j$-th row and $i$-th column of $A^T \pm B^T$ is $a'_{ji} \pm b'_{ji} = a_{ij} \pm b_{ij}$. Since the elements in the $j$-th row and $i$-th column of $(A \pm B)^T$ and $A^T \pm B^T$ are both equal to $a_{ij} \pm b_{ij}$ for all $j=1,\dots,n$ and $i=1,\dots,m$, and both matrices have the same order ($n \times m$), we have $(A \pm B)^T = A^T \pm B^T$.

Property 4: Transpose of a Product (Reversal Law)

The transpose of the product of two matrices is equal to the product of their transposes in the reverse order. This is a very important property and holds true if the matrices $A$ and $B$ have compatible orders for multiplication (i.e., number of columns in $A$ equals number of rows in $B$).

Note the reversal of the order of multiplication on the right side ($B^T$ comes first, then $A^T$). This property extends to the product of multiple matrices: $(ABC)^T = C^T B^T A^T$, and so on.

Derivation: Let $A = [a_{ij}]$ be an $m \times n$ matrix and $B = [b_{jk}]$ be an $n \times p$ matrix. The product $AB = [c_{ik}]$ is an $m \times p$ matrix, where the element in the $i$-th row and $k$-th column is given by the dot product of the $i$-th row of $A$ and the $k$-th column of $B$:

The transpose $(AB)^T = [d_{ki}]$ is a $p \times m$ matrix. The element $d_{ki}$ in the $k$-th row and $i$-th column of $(AB)^T$ is the element in the $i$-th row and $k$-th column of $AB$. So, $d_{ki} = c_{ik} = \sum_{j=1}^n a_{ij} b_{jk}$.

Now consider the product $B^T A^T$. $B^T = [b'_{kj}]$ is a $p \times n$ matrix, where $b'_{kj} = b_{jk}$ (the element in the $k$-th row and $j$-th column of $B^T$ is the element in the $j$-th row and $k$-th column of $B$). $A^T = [a'_{ji}]$ is an $n \times m$ matrix, where $a'_{ji} = a_{ij}$. The product $B^T A^T = [e_{ki}]$ is a $p \times m$ matrix. The element $e_{ki}$ in the $k$-th row and $i$-th column of $B^T A^T$ is the dot product of the $k$-th row of $B^T$ and the $i$-th column of $A^T$.

Substitute the definitions of $(B^T)_{kj}$ and $(A^T)_{ji}$:

So,

Since the order of multiplication within a summation does not matter (scalar multiplication is commutative), we can rewrite the terms:

We see that the element $d_{ki}$ of $(AB)^T$ is equal to the element $e_{ki}$ of $B^T A^T$ for all $k=1,\dots,p$ and $i=1,\dots,m$. Both matrices $(AB)^T$ and $B^T A^T$ are of the same order $p \times m$. Therefore, $(AB)^T = B^T A^T$.

The transpose operation is essential for defining important classes of matrices (like symmetric and skew-symmetric matrices) and for various matrix manipulations and theoretical results in linear algebra.

Symmetric and Skew Symmetric Matrices

Symmetric and skew-symmetric matrices are two important special types of square matrices. Their definitions are based on the relationship between a matrix and its transpose. These types of matrices appear frequently in various areas of mathematics, physics, and engineering due to their unique properties.

Symmetric Matrix

A square matrix $A$ is defined as a symmetric matrix if it is equal to its transpose. That is, when the rows and columns of the matrix are interchanged, the resulting matrix is identical to the original matrix.

If $A = [a_{ij}]$ is an $n \times n$ symmetric matrix, the condition $A^T = A$ implies a specific relationship between its elements. The element in the $i$-th row and $j$-th column of $A$ ($a_{ij}$) must be equal to the element in the $j$-th row and $i$-th column of $A$ ($a_{ji}$) for all possible values of $i$ and $j$ from 1 to $n$. Symbolically: $a_{ij} = a_{ji}$ for all $1 \le i, j \le n$. This means the matrix is symmetric with respect to its main diagonal.

Example of Symmetric Matrix

Consider the matrix $A = \begin{bmatrix} 5 & -1 & 0 \\ -1 & 3 & 7 \\ 0 & 7 & 2 \end{bmatrix}$. This is a square matrix of order $3 \times 3$.

Let's find its transpose $A^T$ by interchanging the rows and columns of $A$.

Comparing the elements of $A^T$ with the elements of $A$, we see that they are identical. Therefore, $A^T = A$, which confirms that $A$ is a symmetric matrix.

Let's verify the element property $a_{ij} = a_{ji}$:

• $a_{12} = -1$ and $a_{21} = -1$. So, $a_{12} = a_{21}$.
• $a_{13} = 0$ and $a_{31} = 0$. So, $a_{13} = a_{31}$.
• $a_{23} = 7$ and $a_{32} = 7$. So, $a_{23} = a_{32}$.

The diagonal elements are $a_{11}=5, a_{22}=3, a_{33}=2$. For diagonal elements, $i=j$, and the condition $a_{ii} = a_{ii}$ is always true.

Skew Symmetric Matrix

A square matrix $A$ is defined as a skew-symmetric matrix (or anti-symmetric matrix) if its transpose is equal to the negative of the matrix itself.

If $A = [a_{ij}]$ is an $n \times n$ skew-symmetric matrix, the condition $A^T = -A$ implies that the element in the $i$-th row and $j$-th column of $A$ ($a_{ij}$) is equal to the negative of the element in the $j$-th row and $i$-th column of $A$ ($a_{ji}$) for all $i, j$. Symbolically: $a_{ij} = -a_{ji}$ for all $1 \le i, j \le n$. This is equivalent to $a_{ij} + a_{ji} = 0$.

Property of Skew Symmetric Matrix: Diagonal Elements are Zero

An important consequence of the definition $a_{ij} = -a_{ji}$ is for the elements on the main diagonal. For these elements, the row index $i$ is equal to the column index $j$ ($i=j$). Applying the condition $a_{ij} = -a_{ji}$ for $i=j$ gives:

Adding $a_{ii}$ to both sides of the equation:

Dividing by 2 (which is a non-zero scalar):

Thus, all the diagonal elements of a skew-symmetric matrix must be zero.

Example of Skew Symmetric Matrix

Consider the matrix $B = \begin{bmatrix} 0 & -4 & 6 \\ 4 & 0 & -8 \\ -6 & 8 & 0 \end{bmatrix}$. This is a square matrix of order $3 \times 3$. Notice the diagonal elements are already zero.

Let's find its transpose $B^T$ by interchanging rows and columns.

Now let's find the negative of matrix $B$, denoted $-B$, by multiplying every element of $B$ by $-1$.

Comparing $B^T$ and $-B$, we see that $B^T = -B$. This confirms that $B$ is a skew-symmetric matrix.

Let's verify the element property $a_{ij} = -a_{ji}$ for off-diagonal elements:

• $a_{12} = -4$ and $a_{21} = 4$. Is $a_{12} = -a_{21}$? Yes, $-4 = -(4)$.
• $a_{13} = 6$ and $a_{31} = -6$. Is $a_{13} = -a_{31}$? Yes, $6 = -(-6)$.
• $a_{23} = -8$ and $a_{32} = 8$. Is $a_{23} = -a_{32}$? Yes, $-8 = -(8)$.

All conditions for a skew-symmetric matrix are satisfied.

Theorem: Representation of a Square Matrix as Sum of Symmetric and Skew Symmetric Matrices

A significant theorem in matrix algebra states that any square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix. This provides a useful way to decompose any square matrix.

Statement of the Theorem: For any square matrix $A$, it can be uniquely written as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$. That is, $A = P + Q$, where $P^T = P$ and $Q^T = -Q$.

Derivation of the Symmetric and Skew-Symmetric Components:

Let $A$ be any square matrix of order $n \times n$. Assume that $A$ can be written as the sum of a symmetric matrix $P$ and a skew-symmetric matrix $Q$:

where $P^T = P$ and $Q^T = -Q$.

Take the transpose of the equation $A = P + Q$. Using the property $(B+C)^T = B^T + C^T$:

Now, substitute the conditions for $P$ being symmetric ($P^T = P$) and $Q$ being skew-symmetric ($Q^T = -Q$) into this equation:

So, we have two matrix equations involving $A, A^T, P,$ and $Q$:

We can solve this system of matrix equations for $P$ and $Q$.

To find $P$, add equation (iii) and equation (iv):

Multiply both sides by the scalar $\frac{1}{2}$:

To find $Q$, subtract equation (iv) from equation (iii):

Multiply both sides by the scalar $\frac{1}{2}$:

These derivations provide explicit formulas for $P$ and $Q$. We can verify that $P = \frac{1}{2}(A+A^T)$ is indeed symmetric and $Q = \frac{1}{2}(A-A^T)$ is indeed skew-symmetric using the properties of transpose (as shown in the previous section):

Also, their sum is $P + Q = \frac{1}{2}(A+A^T) + \frac{1}{2}(A-A^T) = \frac{1}{2}A + \frac{1}{2}A^T + \frac{1}{2}A - \frac{1}{2}A^T = A$.

This shows that any square matrix $A$ can be written as the sum of a symmetric matrix $\underbrace{P = \frac{1}{2}(A+A^T)}_{\text{Symmetric}}$ and a skew-symmetric matrix $\underbrace{Q = \frac{1}{2}(A-A^T)}_{\text{Skew-Symmetric}}$. The uniqueness of this representation is guaranteed because our derivation led to specific formulas for $P$ and $Q$.

Answer:

Symmetric and skew-symmetric matrices are important classes of matrices with distinct properties related to the transpose operation. The ability to decompose any square matrix into the sum of these two types is a powerful result used in various mathematical analyses.

Elementary Operations on a Matrix

Elementary operations, also known as elementary transformations, are a set of basic maneuvers that can be applied to the rows or columns of a matrix. These operations are fundamental in the study of linear algebra and are used extensively for simplifying matrices, solving systems of linear equations, finding the inverse of a matrix, and determining the rank of a matrix.

When an elementary operation is applied to a matrix, the resulting matrix is said to be elementarily equivalent to the original matrix. The key property of elementary operations is that they preserve certain important characteristics of the matrix, such as the set of solutions to a corresponding system of linear equations (if applied as row operations on the augmented matrix) and the rank of the matrix.

Types of Elementary Operations

There are precisely three types of elementary operations. Each type can be applied either to the rows of the matrix or to the columns of the matrix. We will describe each operation by first explaining the row operation and then noting the analogous column operation.

1. Interchange of any two rows (or columns)

This operation involves swapping the positions of two entire rows within the matrix. All the elements in one row are exchanged with the corresponding elements in another row.

Notation: The interchange of the $i$-th row ($R_i$) and the $j$-th row ($R_j$) is denoted by $R_i \leftrightarrow R_j$. The interchange of the $i$-th column ($C_i$) and the $j$-th column ($C_j$) is denoted by $C_i \leftrightarrow C_j$.

Example (Row Operation): Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$. This is a $3 \times 3$ matrix.

Performing the operation $R_1 \leftrightarrow R_3$ means we swap the entire contents of the first row with the entire contents of the third row.

The resulting matrix is elementarily equivalent to $A$. The second row remains in its original position.

2. Multiplication of the elements of any row (or column) by a non-zero scalar

This operation involves choosing a specific row (or column) and multiplying each and every element in that row (or column) by the same non-zero real number (scalar). The scalar must be non-zero because multiplying a row by zero would effectively remove all information from that row, which is not a reversible process (you cannot uniquely recover the original row from a row of zeros), thus violating the property of elementary equivalence needed for operations like finding the inverse.

Notation: The multiplication of the $i$-th row ($R_i$) by a non-zero scalar $k$ is denoted by $R_i \to k R_i$. The multiplication of the $i$-th column ($C_i$) by a non-zero scalar $k$ is denoted by $C_i \to k C_i$. The condition $k \neq 0$ is essential.

Example (Row Operation): Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$. Performing the operation $R_2 \to 5 R_2$ means we multiply every element in the second row by the scalar 5.

The first and third rows are unchanged. Each element in the second row is five times its original value.

3. Addition to the elements of any row (or column), the corresponding elements of any other row (or column) multiplied by a non-zero scalar

This is the most powerful elementary operation for transforming matrices. It involves selecting a "target" row ($R_i$) and a "source" row ($R_j$). We multiply every element of the source row $R_j$ by a non-zero scalar $k$, and then add the resulting values to the corresponding elements of the target row $R_i$. The source row $R_j$ remains unchanged; only the target row $R_i$ is modified. This operation is typically used to create zeros in specific positions within the matrix.

Notation: The operation of adding $k$ times the $j$-th row ($R_j$) to the $i$-th row ($R_i$) is denoted by $R_i \to R_i + k R_j$. The analogous column operation is $C_i \to C_i + k C_j$. It is essential that $k$ is a non-zero scalar ($k \neq 0$) and that $i \neq j$ (since adding a multiple of a row to itself is covered by operation 2, unless $k=-1$, which would be $R_i \to R_i - R_i = R_i \to 0$, also covered by operation 2 with $k=0$, which is not allowed).

Example (Row Operation): Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$. Performing the operation $R_2 \to R_2 + 2 R_1$ means we multiply the first row ($R_1$) by 2 and add the result to the second row ($R_2$).

Original $R_2 = [4 \ 5 \ 6]$. $2 R_1 = [2 \times 1 \ 2 \times 2 \ 2 \times 3] = [2 \ 4 \ 6]$.

New $R_2 = [4+2 \ 5+4 \ 6+6] = [6 \ 9 \ 12]$.

The first and third rows remain unchanged. The second row is modified by adding twice the corresponding elements from the first row.

Important Note on Usage: When solving systems of linear equations using augmented matrices or finding the inverse of a matrix by reducing the augmented matrix $[A | I]$ to $[I | A^{-1}]$, it is crucial to perform either only row operations throughout the entire process or only column operations throughout the entire process. Applying a mix of row and column operations simultaneously on an augmented matrix will generally lead to an incorrect result for the inverse matrix or the solution of the system in that method.

Elementary operations are the backbone of many matrix manipulation techniques and algorithms in linear algebra.

Invertible Matrices (Definition)

In the realm of real numbers, every non-zero number $a$ has a multiplicative inverse, its reciprocal $1/a$, such that $a \times (1/a) = 1$. The concept of an inverse extends to square matrices, providing a way to define matrix "division" in certain cases. A square matrix that behaves analogously to a non-zero number in terms of multiplication is called an invertible matrix.

Definition of an Invertible Matrix

A square matrix $A$ of order $n \times n$ is said to be invertible or non-singular if there exists another square matrix $B$ of the exact same order $n \times n$ such that the product of $A$ and $B$, in both possible orders of multiplication, results in the identity matrix of order $n$, denoted by $I_n$.

If such a matrix $B$ exists, it is called the inverse of matrix $A$. A significant property is that if an inverse exists, it is unique (as proven below). The unique inverse of matrix $A$ is denoted by the symbol $A^{-1}$. Thus, if $A$ is an invertible matrix, its inverse $A^{-1}$ satisfies the condition:

A square matrix that is not invertible is called a singular matrix. Singular matrices do not have an inverse satisfying the definition above.

Important Note: The concept of an inverse matrix as defined here ($A A^{-1} = A^{-1}A = I_n$) is strictly applicable only to square matrices. Non-square matrices do not have two-sided inverses in this sense, although related concepts like left inverses ($BA = I_n$) or right inverses ($AB = I_n$) may exist for some rectangular matrices, where $B$ and $A$ might have different dimensions, and $I_n$ and $I_m$ would be identity matrices of different orders.

Uniqueness of the Inverse

A crucial property of the inverse matrix is that for a given square matrix, its inverse (if it exists) is one-of-a-kind. There cannot be two different matrices that both satisfy the definition of being the inverse of the same matrix.

Proof of Uniqueness:

Given: Let $A$ be a square matrix of order $n$. Assume that $A$ is invertible, and suppose, for the sake of contradiction, that there exist two different matrices, $B$ and $C$, both of order $n$, that are inverses of $A$.

To Prove: The assumption that $B$ and $C$ are different must lead to a contradiction, thereby proving that $B$ and $C$ must be the same matrix. We need to prove that $B = C$.

Proof:

Since $B$ is assumed to be an inverse of $A$, by the definition of an invertible matrix, we have:

Since $C$ is also assumed to be an inverse of $A$, by the definition, we have:

We want to show that matrix $B$ is identical to matrix $C$. Let's start with the expression $B I_n$. We know from the properties of matrix multiplication that multiplying any matrix by the identity matrix of the compatible size results in the original matrix. Since $B$ is an $n \times n$ matrix and $I_n$ is the identity matrix of order $n$, we have $B I_n = B$.

From equation (iii), we know that $I_n = AC$. Substitute this expression for $I_n$ into the equation above:

Matrix multiplication is associative, meaning the way matrices are grouped for multiplication does not affect the final product, provided the order of the matrices is maintained. Apply the associative property to the right side $B(AC)$:

From equation (ii), we know that $BA = I_n$. Substitute this expression for $BA$ into the equation:

Again, using the property of the identity matrix, multiplying $I_n$ by any matrix $C$ results in $C$ (since $C$ is $n \times n$ and $I_n$ is order $n$): $I_n C = C$. Substitute this into the equation:

This result shows that the two matrices, $B$ and $C$, which were assumed to be different inverses of $A$, must in fact be equal. This contradicts our initial assumption that $B$ and $C$ were different. Therefore, the initial assumption must be false. This proves that if a matrix $A$ has an inverse, that inverse is unique.

Property: Inverse of a Product of Invertible Matrices

If we have two or more square matrices that are all invertible and have the same order, their product is also invertible. Furthermore, there is a specific rule for determining the inverse of such a product.

Statement: If $A$ and $B$ are invertible matrices of the same order $n \times n$, then their product $AB$ is also an invertible matrix, and its inverse is given by the product of the inverses of $B$ and $A$ taken in the reverse order:

This property is known as the Reversal Law for Inverses and is analogous to the reversal law for the transpose of a product ($(AB)^T = B^T A^T$). This property extends to any finite number of invertible matrices of the same order, e.g., $(ABC)^{-1} = C^{-1} B^{-1} A^{-1}$.

Proof of the Reversal Law for Inverses:

Given: Let $A$ and $B$ be invertible matrices of order $n$. This means their inverses, $A^{-1}$ and $B^{-1}$, exist and are also of order $n$. By the definition of the inverse, they satisfy $AA^{-1} = A^{-1}A = I_n$ and $BB^{-1} = B^{-1}B = I_n$. The product matrix $AB$ is defined and is a square matrix of order $n \times n$.

To Prove: The matrix $B^{-1}A^{-1}$ is the inverse of the matrix $AB$. According to the definition of an inverse, we must show that when the matrix $AB$ is multiplied by the matrix $B^{-1}A^{-1}$ (in both orders), the result is the identity matrix $I_n$. That is, we need to prove that $(AB)(B^{-1}A^{-1}) = I_n$ and $(B^{-1}A^{-1})(AB) = I_n$.

Consider the first product: $(AB)(B^{-1}A^{-1})$

Using the associative property of matrix multiplication, we can group the matrices differently:

From the definition of the inverse (given), we know that $B B^{-1} = I_n$. Substitute the identity matrix $I_n$ into the expression:

We know from the properties of the identity matrix that multiplying any matrix by the identity matrix (of the compatible size) leaves the matrix unchanged. Here, $A I_n = A$. Substitute this back:

Finally, from the definition of the inverse (given), we know that $A A^{-1} = I_n$.

Thus, we have shown that $(AB)(B^{-1}A^{-1}) = I_n$.

Now, consider the second product: $(B^{-1}A^{-1})(AB)$

Using the associative property of matrix multiplication, we can group the matrices differently:

From the definition of the inverse (given), we know that $A^{-1} A = I_n$. Substitute the identity matrix $I_n$ into the expression:

We know from the properties of the identity matrix that multiplying the identity matrix $I_n$ by any matrix $B$ leaves the matrix unchanged ($I_n B = B$). Substitute this back:

Finally, from the definition of the inverse (given), we know that $B^{-1} B = I_n$.

So, we have also shown that $(B^{-1}A^{-1})(AB) = I_n$.

Since multiplying $AB$ by $B^{-1}A^{-1}$ in both orders results in the identity matrix $I_n$, by the definition of the inverse, the matrix $B^{-1}A^{-1}$ is indeed the inverse of the matrix $AB$.

This completes the proof of the reversal law for the inverse of a product of invertible matrices.

The concept of invertible matrices is fundamental for solving matrix equations, representing inverse transformations, and plays a critical role in areas like linear systems and determinants.