Question

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

Hint:

This property is often called the "socks and shoes" rule. To undo the process of putting on socks then shoes, you must first take off the shoes and then take off the socks (the reverse order). Similarly, the inverse of a product of matrices is the product of their inverses in the reverse order.

Answer 1

Step 1: Understanding the Concept:
We need to prove the reversal law for the inverse of a product of matrices. The proof relies on the fundamental definition of a matrix inverse: a matrix M is the inverse of N if their product, in either order, is the identity matrix I (i.e., \( MN = NM = I \)).
Step 2: Key Formula or Approach:
To prove that \( B^{-1}A^{-1} \) is the inverse of \( (AB) \), we need to show that:
1. \( (AB)(B^{-1}A^{-1}) = I \)
2. \( (B^{-1}A^{-1})(AB) = I \)
Step 3: Detailed Explanation or Calculation:
Let's start with the product \( (AB) \) and \( (B^{-1}A^{-1}) \).
Proof of the first condition:
\[ (AB)(B^{-1}A^{-1}) \] 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, \( BB^{-1} = I \) (the identity matrix).
\[ = A(I)A^{-1} \] The product of any matrix with the identity matrix is the matrix itself (\( AI = A \)).
\[ = AA^{-1} \] Since A is an invertible matrix, by definition, \( AA^{-1} = I \).
\[ = I \] So, we have shown that \( (AB)(B^{-1}A^{-1}) = I \).
Proof of the second condition:
Now, let's check the product in the reverse order:
\[ (B^{-1}A^{-1})(AB) \] Using the associative property:
\[ = B^{-1}(A^{-1}A)B \] Since A is invertible, \( A^{-1}A = I \).
\[ = B^{-1}(I)B \] Since \( B^{-1}I = B^{-1} \):
\[ = B^{-1}B \] Since B is invertible, \( B^{-1}B = I \).
\[ = I \] So, we have also shown that \( (B^{-1}A^{-1})(AB) = I \).
Step 4: Final Answer:
Since \( (AB)(B^{-1}A^{-1}) = I \) and \( (B^{-1}A^{-1})(AB) = I \), by the definition of an inverse, the inverse of the matrix \( (AB) \) is \( B^{-1}A^{-1} \). Hence, \( (AB)^{-1} = B^{-1}A^{-1} \) is proved.