If \(A = \begin{bmatrix}2 & 3\\1 & -4\end{bmatrix}\) and \(B = \begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}\), then \(B^{-1}A^{-1}\) is equal to:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{1}{11}\begin{bmatrix}14 & 5\\5 & 1\end{bmatrix}\)
By the reversal law, \((AB)^{-1} = B^{-1}A^{-1}\), so only one inverse is needed:
\( \displaystyle AB = \begin{bmatrix}2 - 3 & -4 + 9\\1 + 4 & -2 - 12\end{bmatrix} = \begin{bmatrix}-1 & 5\\5 & -14\end{bmatrix} \)
\(|AB| = (-1)(-14) - 5\times 5 = 14 - 25 = -11\)
For a 2 × 2 matrix, the adjoint swaps the diagonal entries and changes the signs of the other two:
\( \displaystyle (AB)^{-1} = \frac{1}{-11}\begin{bmatrix}-14 & -5\\-5 & -1\end{bmatrix} \)
\( \displaystyle = \frac{1}{11}\begin{bmatrix}14 & 5\\5 & 1\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(-\frac{1}{11}\begin{bmatrix}14 & 5\\5 & 1\end{bmatrix}\) — Incorrect. This is the answer with the wrong sign; both |AB| = −11 and every adjoint entry are negative, so the signs cancel.
Option (2) \(\frac{1}{11}\begin{bmatrix}15 & 11\\1 & 0\end{bmatrix}\) — Incorrect. This is \((BA)^{-1} = A^{-1}B^{-1}\), the inverses taken in the wrong order. Here \(BA = \begin{bmatrix}0 & 11\\1 & -15\end{bmatrix}\), with |BA| = −11.
Option (4) \(-\frac{1}{11}\begin{bmatrix}15 & 11\\1 & 0\end{bmatrix}\) — Incorrect. This is \(A^{-1}B^{-1}\) (wrong order) with the wrong sign as well.