If \(A =\begin{bmatrix}3&-2&3\\2&1&-1\\4&-3&2\end{bmatrix}\), then the matrix \((\mathrm{adj}\, A)A\) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(-17\, I\)
For every square matrix, \((\mathrm{adj}\, A)A = A(\mathrm{adj}\, A) = |A|\,I\). So only the determinant is needed. Expanding along the first row:
\(\displaystyle |A| = 3\begin{vmatrix}1&-1\\-3&2\end{vmatrix} + 2\begin{vmatrix}2&-1\\4&2\end{vmatrix} + 3\begin{vmatrix}2&1\\4&-3\end{vmatrix}\)
\(= 3(2 - 3) + 2(4 + 4) + 3(-6 - 4)\)
\(= -3 + 16 - 30 = -17\)
\((\mathrm{adj}\, A)A = -17\, I\)