If A is a 3 × 3 matrix such that \(|\mathrm{adj}\,A| = 9\) and \(|kA^{-1}| = 9\), then the value of \(k\) are:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\pm 3\)
For a matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\) and \(|kB| = k^n|B|\). Here n = 3.
From the determinant of the adjoint:
\(|A|^2 = 9,\quad |A| = \pm 3\)
Since \(|A^{-1}| = \frac{1}{|A|}\):
\( \displaystyle |kA^{-1}| = k^3\,|A^{-1}| = \frac{k^3}{|A|} \)
\( \displaystyle \frac{k^3}{|A|} = 9 \)
\(k^3 = 9|A| = \pm 27\)
\(k = \pm 3\)
(k = 3 when |A| = 3, and k = −3 when |A| = −3.)