If \(A=\begin{bmatrix}1&2&1\\2&3&1\\0&0&1\end{bmatrix}\) then \(|\mathrm{adj}\,(3A^T)|^2\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(3^{12}\)
Expanding \(|A|\) along the third row, which has two zeros:
\( |A| = 1 \times \begin{vmatrix}1&2\\2&3\end{vmatrix} \)
\(|A| = 3 - 4 = -1\)
For a square matrix M of order n, \(|kM| = k^n|M|\), \(|M^T| = |M|\) and \(|\mathrm{adj}\,M| = |M|^{n-1}\). Here n = 3. Let \(M = 3A^T\):
\(|M| = 3^3\,|A^T| = 27 \times (-1) = -27\)
\(|\mathrm{adj}\,M| = |M|^2 = (-27)^2 = 3^6\)
\(|\mathrm{adj}\,(3A^T)|^2 = (3^6)^2 = 3^{12}\)