If \(A = \begin{bmatrix}0&0&\sqrt{3}\\0&\sqrt{3}&0\\\sqrt{3}&0&0\end{bmatrix}\), then \(|\mathrm{adj}\, A|\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 27
For a square matrix of order n, the determinant of the adjoint is \(|\mathrm{adj}\,A| = |A|^{n-1}\). Here n = 3, so \(|\mathrm{adj}\,A| = |A|^2\).
Expanding |A| along the first row, only the entry \(\sqrt{3}\) in position (1, 3) contributes:
\( \displaystyle |A| = \sqrt{3}\begin{vmatrix}0&\sqrt{3}\\\sqrt{3}&0\end{vmatrix} \)
\(|A| = \sqrt{3}\,(0 - 3) = -3\sqrt{3}\)
Hence:
\(|\mathrm{adj}\,A| = (-3\sqrt{3})^2 = 27\)