Question

For two invertible matrices A and B of suitable orders, the value of \((AB)^{-1}\) is:

• A. \((BA)^{-1}\)
• B. \(B^{-1}A^{-1}\)
• C. \(A^{-1}B^{-1}\)
• D. \((AB')^{-1}\)

Hint:

The inverse of a product reverses the order of the factors.

Answer 1

The Correct Option is B

Step 1: Recall the reversal rule for the inverse of a product of matrices: for two invertible matrices \(A\) and \(B\) of compatible orders, \((AB)^{-1} = B^{-1}A^{-1}\).
Step 2: This can be verified directly. We want \(X\) such that \((AB)X = I\). Try \(X = B^{-1}A^{-1}\): \((AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = A(I)A^{-1} = AA^{-1} = I\). Similarly \((B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}(I)B = B^{-1}B = I\). Since X satisfies both conditions, it is the unique inverse of AB.
Step 3: Option 1, \((BA)^{-1} = A^{-1}B^{-1}\), is the inverse of BA, not of AB, and these are generally different since matrix multiplication is not commutative. Option 3 fails since \((AB)(A^{-1}B^{-1}) = A(BA^{-1})B^{-1}\), which does not simplify to I unless A and B commute. Option 4 introduces a transpose \(B'\) that was never part of the original product.
Step 4: Hence the correct relation is \((AB)^{-1} = B^{-1}A^{-1}\), option 2.