If \(A = \begin{bmatrix}1 & 2\\ 3 & -1\end{bmatrix}\) and \(A^{-1} = kA\), then the value of k is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle \frac{1}{7}\)
The determinant and the adjoint of A:
\( |A| = 1 \times (-1) - 2 \times 3 = -7 \)
\( \displaystyle \mathrm{adj}\,A = \begin{bmatrix}-1 & -2\\ -3 & 1\end{bmatrix} = -A \)
\( \displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A = \frac{-A}{-7} = \frac{1}{7}A \)
So \(\displaystyle k = \frac{1}{7}\).
Check: \(A^2 = \begin{bmatrix}1 + 6 & 2 - 2\\ 3 - 3 & 6 + 1\end{bmatrix} = 7I\), so \(A \times \frac{1}{7}A = I\).