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Chapter 3: Problem 75

If \(A\) and \(B\) are invertible, check that \(B^{-1} A^{-1}\) is the inverse of \(A B\).

Short Answer

Expert verified
We need to prove that \(B^{-1}A^{-1}\) is the inverse of \(AB\). Using the associative property of matrix multiplication and the property of invertible matrices, we calculate the product of the two matrices in both orders and obtain: \( (B^{-1} A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}IB = B^{-1}B = I \) \( (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I \) Thus, we have shown that \(B^{-1}A^{-1}\) is indeed the inverse of \(AB\).

Step by step solution

01

Compute the product of (B^{-1} A^{-1})(AB) and (AB)(B^{-1}A^{-1})

Let's compute the product of the two matrices (B^{-1} A^{-1})(AB) and (AB)(B^{-1}A^{-1}) separately.
02

Compute (B^{-1} A^{-1})(AB)

To compute (B^{-1} A^{-1})(AB), we will use the associative property of matrix multiplication which states that (AB)C = A(BC) for any matrices A, B and C, so: (B^{-1} A^{-1})(AB) = B^{-1}(A^{-1}A)B Now, since A is invertible, it means that A^{-1}A = I, where I is the identity matrix, so: B^{-1}(A^{-1}A)B = B^{-1}IB Since the identity matrix I does not affect the product when multiplied with a matrix, we are left with: B^{-1}B Now, as B is invertible, it means B^{-1}B = I, so: (B^{-1} A^{-1})(AB) = I
03

Compute (AB)(B^{-1}A^{-1})

To compute (AB)(B^{-1}A^{-1}), we will use the associative property of matrix multiplication again: (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} Since B is invertible, BB^{-1} = I, so: A(BB^{-1})A^{-1} = AIA^{-1} Now, since the identity matrix I does not affect the product when multiplied with a matrix, we are left with: AA^{-1} And as A is invertible, it means AA^{-1} = I, so: (AB)(B^{-1}A^{-1}) = I
04

Conclusion

Based on the previous steps, we showed that: (B^{-1} A^{-1})(AB) = I and (AB)(B^{-1}A^{-1}) = I This proves that B^{-1}A^{-1} is indeed the inverse of AB.

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