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The identity matrix, often denoted as I, is a special kind of square matrix. It plays a role similar to the number 1 in real number multiplication. When any matrix is multiplied by the identity matrix, it remains unchanged. This is what we mean by the properties:

• IA = A
• AI = A

Here, A can be any n x n matrix. The identity matrix has 1’s along the main diagonal (from the top left to the bottom right) and all other entries are 0. For example, the 2x2 identity matrix is:
\[ I_{2} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} \] Similarly, for a 3x3 matrix:
\[ I_{3} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \] This is an essential component in understanding the solution, as the identity matrix serves as a neutral element in matrix multiplication.

Matrix multiplication involves combining two matrices to yield a third matrix. The process is not as simple as multiplying each corresponding element; instead, it consists of the dot product of rows and columns. If A is an m x n matrix and B is an n x p matrix, their product AB will be an m x p matrix.
For two matrices A and B to be multiplied together, the number of columns in A must be equal to the number of rows in B. This step is crucial to ensure compatibility for multiplication.
The element in the i-th row and j-th column of the resulting matrix is computed as follows:
\[ (AB)_{ij} = \sum_{k=1}^n A_{ik} B_{kj} \]

• This means you multiply each element of the i-th row of A by the corresponding element of the j-th column of B and then sum those products.
• Matrix multiplication is associative but not commutative. This means, in general, \[ AB eq BA \].

In our given problem, we know that AB = I. This information, coupled with matrix properties, will help us demonstrate that BA = I.

An inverse matrix of a square matrix A is another matrix, denoted as A^{-1}, such that their product is the identity matrix. Simply put, for a matrix A:
\[ A A^{-1} = A^{-1} A = I \] A matrix A must be square (having the same number of rows and columns) to have an inverse. In the given exercise, if AB = I, then A and B are inverses of each other. This can be understood as follows:

• If A can be multiplied by B to yield the identity matrix I, then A is the inverse of B, and vice versa.
• This property is expressed as \[ B = A^{-1} \text{ and } A = B^{-1} \].

Using this property, we can rewrite B as A^{-1} and thus verify that:
\[ B A = A^{-1} A = I \] This consolidates that if the product of matrices A and B equals the identity matrix, then the product of B and A also equals the identity matrix. This is the core justification behind the exercise.