If \(A = \begin{bmatrix}7&3\\5&-7\end{bmatrix}\) be such that \(A^{-1} = kA\), then \(k\) equals
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{1}{64}\)
For a 2 × 2 matrix, \(A^{-1} = \displaystyle\frac{1}{|A|}\,\mathrm{adj}\,A\).
\(|A| = 7(-7) - 3(5) = -49 - 15 = -64\)
The adjoint swaps the diagonal entries and changes the signs of the other two:
\( \displaystyle \mathrm{adj}\,A = \begin{bmatrix}-7&-3\\-5&7\end{bmatrix} = -A \)
Hence
\( \displaystyle A^{-1} = \frac{-A}{-64} = \frac{1}{64}A \)
so \(k = \displaystyle\frac{1}{64}\).