If \(A = \begin{bmatrix}1 & 2\\ 1 & 3\end{bmatrix}\) and \(B^{-1} = \begin{bmatrix}2 & 1\\ 1 & 1\end{bmatrix}\), then \((AB)^{-1}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\begin{bmatrix}5 & -3\\ 2 & -1\end{bmatrix}\)
The inverse of a product reverses the order: \((AB)^{-1} = B^{-1}A^{-1}\).
For A:
\(|A| = 1 \cdot 3 - 2 \cdot 1 = 1\)
\( \displaystyle A^{-1} = \frac{1}{1}\begin{bmatrix}3 & -2\\ -1 & 1\end{bmatrix} \)
Then
\( \displaystyle B^{-1}A^{-1} = \begin{bmatrix}2 & 1\\ 1 & 1\end{bmatrix}\begin{bmatrix}3 & -2\\ -1 & 1\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}6 - 1 & -4 + 1\\ 3 - 1 & -2 + 1\end{bmatrix} = \begin{bmatrix}5 & -3\\ 2 & -1\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}4 & 1\\ -1 & 0\end{bmatrix}\) — Incorrect. This is \(A^{-1}B^{-1}\), the product in the wrong order.
Option (2) \(\begin{bmatrix}5 & 2\\ -3 & -1\end{bmatrix}\) — Incorrect. This is the transpose of \(B^{-1}A^{-1}\).
Option (4) \(\begin{bmatrix}-1 & 3\\ -2 & 5\end{bmatrix}\) — Incorrect. This is the adjoint of the correct answer, which is the matrix AB itself, not its inverse.