If \(A = \begin{bmatrix} 1 & 3 & 5\\ 1 & 0 & 3\\ 0 & 1 & 0\end{bmatrix}\), then \(|(\mathrm{adj}A)|\) is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
For a square matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\). Here n = 3, so only |A| is needed.
Expanding along row 3, which has two zeros (the entry 1 is in position (3, 2), sign −):
\( |A| = -1 \times \begin{vmatrix} 1 & 5\\ 1 & 3\end{vmatrix} \)
\( = -1 \times (3 - 5) = 2 \)
Then
\(|\mathrm{adj}\,A| = |A|^{3-1} = 2^2 = 4\)
Why the other options are incorrect
Option (1) 2 — Incorrect. 2 is |A| itself, not |adj A|.
Option (2) 8 — Incorrect. 8 is \(|A|^3\); the power in \(|\mathrm{adj}\,A| = |A|^{n-1}\) is n − 1 = 2, not n = 3.
Option (4) 25 — Incorrect. With |A| = 2, \(|\mathrm{adj}\,A| = 4\), not 25.