If \(A = \begin{bmatrix}3 & 1\\ 4 & 2\end{bmatrix}\), then \((\mathrm{adj}\,A)^{-1}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle \frac{1}{2}\begin{bmatrix}3 & 1\\ 4 & 2\end{bmatrix}\)
Since \(A(\mathrm{adj}\,A) = |A|\,I\), we have \(\displaystyle (\mathrm{adj}\,A)\cdot\frac{A}{|A|} = I\), so the inverse of the adjoint is:
\(\displaystyle (\mathrm{adj}\,A)^{-1} = \frac{1}{|A|}A\)
\(|A| = 3 \cdot 2 - 1 \cdot 4 = 2\)
\(\displaystyle (\mathrm{adj}\,A)^{-1} = \frac{1}{2}\begin{bmatrix}3 & 1\\ 4 & 2\end{bmatrix}\)
Check: \(\displaystyle \mathrm{adj}\,A = \begin{bmatrix}2 & -1\\ -4 & 3\end{bmatrix}\), and \(\displaystyle \frac{1}{2}\begin{bmatrix}2 & -1\\ -4 & 3\end{bmatrix}\begin{bmatrix}3 & 1\\ 4 & 2\end{bmatrix} = \frac{1}{2}\begin{bmatrix}2 & 0\\ 0 & 2\end{bmatrix} = I\).
Why the other options are incorrect
Option (1) \(\begin{bmatrix}2 & -1\\ -4 & 3\end{bmatrix}\) — Incorrect. This is adj A itself, not its inverse.
Option (2) \(\displaystyle \frac{1}{2}\begin{bmatrix}2 & -1\\ -4 & 3\end{bmatrix}\) — Incorrect. This is \(A^{-1}\), the inverse of A, not of adj A.
Option (4) \(2\begin{bmatrix}3 & 1\\ 4 & 2\end{bmatrix}\) — Incorrect. This is \(|A|\,A\); A has to be divided by \(|A|\), not multiplied.