If $A =\begin{bmatrix}3&-2&3\\2&1&-1\\4&-3&2\end{bmatrix}$, then the matrix $\text{(adj A)A}$ is equal to

-17 I

The correct answer is Option (2) → -17 I

Given: $A = \begin{bmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{bmatrix}$

Property: $(\text{adj}\ A)\cdot A = |A| \cdot I$

Compute $|A|$:

$|A| = \begin{vmatrix} 3 & -2 & 3 \\ 2 & 1 & -1 \\ 4 & -3 & 2 \end{vmatrix}$

= $3 \begin{vmatrix} 1 & -1 \\ -3 & 2 \end{vmatrix} - (-2) \begin{vmatrix} 2 & -1 \\ 4 & 2 \end{vmatrix} + 3 \begin{vmatrix} 2 & 1 \\ 4 & -3 \end{vmatrix}$

= $3(1 \cdot 2 - (-1)(-3)) + 2(2 \cdot 2 - (-1)(4)) + 3(2 \cdot -3 - 1 \cdot 4)$

= $3(2 - 3) + 2(4 + 4) + 3(-6 - 4)$

= $3(-1) + 2(8) + 3(-10) = -3 + 16 - 30 = -17$

Then:

$(\text{adj}\ A)\cdot A = |A| \cdot I = -17 \cdot I = -17 \cdot \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} -17 & 0 & 0 \\ 0 & -17 & 0 \\ 0 & 0 & -17 \end{bmatrix}$