Let A and B be invertible matrices of order 2 with \(A = \begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix}\) and \(B^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\). Then \((A^{-1}B)^{-1}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\)
By the reversal law \((PQ)^{-1} = Q^{-1}P^{-1}\), and since \((A^{-1})^{-1} = A\):
\((A^{-1}B)^{-1} = B^{-1}(A^{-1})^{-1} = B^{-1}A\)
So no inverse has to be worked out; only one product is needed:
\( \displaystyle B^{-1}A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 2 \\ 1 & 3 \end{bmatrix} \)
\( \displaystyle = \begin{bmatrix} 2 + 1 & 4 + 3 \\ 1 + 1 & 2 + 3 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix} 4 & 3 \\ 5 & 4 \end{bmatrix}\) — Incorrect. This is \(AB^{-1}\); the order of the product has not been reversed.
Option (3) \(\begin{bmatrix} 3 & 2 \\ 7 & 5 \end{bmatrix}\) — Incorrect. This is the transpose of \(B^{-1}A\).
Option (4) \(\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}\) — Incorrect. This is \(A^{-1}B\) itself, the inverse of the required matrix.