If \(A = \begin{bmatrix}2 & 1 & 0\\ 0 & 3 & 1\\ 1 & 0 & 1\end{bmatrix}\), then \(|\mathrm{adj}\,A|\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 49
Expanding along the first row:
\(|A| = 2(3 \cdot 1 - 1 \cdot 0) - 1(0 \cdot 1 - 1 \cdot 1) + 0\)
\(|A| = 6 + 1\)
\(|A| = 7\)
For a square matrix of order n, \(|\mathrm{adj}\,A| = |A|^{n-1}\). Here n = 3:
\(|\mathrm{adj}\,A| = 7^{2}\)
\(|\mathrm{adj}\,A| = 49\)
Why the other options are incorrect
Option (1) 7 — Incorrect. This is \(|A|\) itself, not \(|\mathrm{adj}\,A|\).
Option (2) 14 — Incorrect. This is \(2|A|\); the result is a power of \(|A|\), not a multiple.
Option (3) 343 — Incorrect. This is \(7^{3}\); the power is \(n - 1 = 2\), not n.