If \(A = \begin{bmatrix}4&3\\2&2\end{bmatrix}\), then \(A^{-1}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}1&-\displaystyle\frac{3}{2}\\-1&2\end{bmatrix}\)
For \(A = \begin{bmatrix}a&b\\c&d\end{bmatrix}\) with \(ad - bc \ne 0\), the inverse is obtained by swapping a and d, changing the signs of b and c, and dividing by \(ad - bc\).
\(ad - bc = 4(2) - 3(2) = 2\)
\( \displaystyle A^{-1} = \frac{1}{2}\begin{bmatrix}2&-3\\-2&4\end{bmatrix} \)
\( \displaystyle = \begin{bmatrix}1&-\frac{3}{2}\\-1&2\end{bmatrix} \)
Check: the first row of A times the first column of \(A^{-1}\) is \(4(1) + 3(-1) = 1\), and the first row times the second column is \(4\left(-\frac{3}{2}\right) + 3(2) = 0\).
Why the other options are incorrect
Option (1) \(\begin{bmatrix}2&-3\\-2&4\end{bmatrix}\) — Incorrect. This is the adjoint of A; it has not been divided by \(ad - bc = 2\).
Option (2) \(\begin{bmatrix}1&\displaystyle\frac{3}{2}\\1&2\end{bmatrix}\) — Incorrect. The signs of b and c were not changed.
Option (3) \(\begin{bmatrix}2&-\displaystyle\frac{3}{2}\\-1&1\end{bmatrix}\) — Incorrect. The diagonal elements 4 and 2 were not swapped.