If \(A = \begin{bmatrix}1&2\\0&1\end{bmatrix}\) and \(B = \begin{bmatrix}1&0\\3&1\end{bmatrix}\), then \((AB)^{-1}\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\begin{bmatrix}1&-2\\-3&7\end{bmatrix}\)
The product, row by column:
\( \displaystyle AB = \begin{bmatrix}1 + 6&0 + 2\\0 + 3&0 + 1\end{bmatrix} = \begin{bmatrix}7&2\\3&1\end{bmatrix} \)
For this matrix \(ad - bc = 7 - 6 = 1\), so swapping the diagonal elements and changing the signs of the other two:
\( \displaystyle (AB)^{-1} = \begin{bmatrix}1&-2\\-3&7\end{bmatrix} \)
This agrees with the reversal rule \((AB)^{-1} = B^{-1}A^{-1}\):
\( \displaystyle B^{-1}A^{-1} = \begin{bmatrix}1&0\\-3&1\end{bmatrix}\begin{bmatrix}1&-2\\0&1\end{bmatrix} = \begin{bmatrix}1&-2\\-3&7\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}7&-2\\-3&1\end{bmatrix}\) — Incorrect. This is \(A^{-1}B^{-1}\), the inverses multiplied in the wrong order; it is the inverse of BA, not of AB.
Option (3) \(\begin{bmatrix}7&2\\3&1\end{bmatrix}\) — Incorrect. This is AB itself.
Option (4) \(\begin{bmatrix}1&2\\3&7\end{bmatrix}\) — Incorrect. This is BA; its product with AB is not I (its first element is \(7 + 6 = 13\)).