Question

If \( A \) and \( B \) are two matrices of order \( n \) which are invertible, then prove that \( (AB)^{-1} = B^{-1}A^{-1} \).

Hint:

This is often called the "reversal rule" for inverses (or the "socks and shoes" property: to undo putting on socks then shoes, you must first take off the shoes, then the socks). The same reversal rule applies to the transpose of a product of matrices: \( (AB)^T = B^T A^T \).

Answer 1

Step 1: Understanding the Concept:
The inverse of a square matrix M is a matrix \( M^{-1} \) such that their product is the identity matrix I, i.e., \( M M^{-1} = M^{-1} M = I \). To prove the given identity, we must show that multiplying \( AB \) by \( B^{-1}A^{-1} \) results in the identity matrix.
Step 2: Key Approach:
We will use the definition of an inverse. We will show that \( (AB)(B^{-1}A^{-1}) = I \).
Step 3: Detailed Proof:
Consider the product of the matrix \( (AB) \) and the matrix \( (B^{-1}A^{-1}) \): \[ (AB)(B^{-1}A^{-1}) \] Using the associative property of matrix multiplication, we can regroup the terms: \[ = A(BB^{-1})A^{-1} \] Since B is an invertible matrix, by definition, \( BB^{-1} = I \), where I is the identity matrix of order n. \[ = A(I)A^{-1} \] The product of any matrix and the identity matrix is the matrix itself, so \( AI = A \). \[ = AA^{-1} \] Since A is an invertible matrix, by definition, \( AA^{-1} = I \). \[ = I \] Thus, we have shown that \( (AB)(B^{-1}A^{-1}) = I \). This proves that \( B^{-1}A^{-1} \) is the right inverse of \( AB \). For square matrices, a right inverse is also a left inverse, so \( B^{-1}A^{-1} \) is the inverse of \( AB \).
Step 4: Final Answer:
By definition of a matrix inverse and the associative property of multiplication, we have proved that \( (AB)^{-1} = B^{-1}A^{-1} \).