If \(A = \begin{bmatrix}0&0&\sqrt{7}\\0&\sqrt{7}&0\\\sqrt{7}&0&0\end{bmatrix}\), then \(|\mathrm{adj}\,A|\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 343
Expanding \(|A|\) along the first row, only the third entry is non-zero:
\( \displaystyle |A| = \sqrt{7}\begin{vmatrix}0&\sqrt{7}\\\sqrt{7}&0\end{vmatrix} \)
\(= \sqrt{7}\,(0 - 7) = -7\sqrt{7}\)
For a square matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\). Here \(n = 3\):
\(|\mathrm{adj}\,A| = (-7\sqrt{7})^2\)
\(= 49 \times 7 = 343\)