Let $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$. Verify that $A(\text{adj } A) = (\text{adj } A) A = |A| I$.

(A) Given $A = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix}$.
First,calculate the determinant $|A|$:
$|A| = (2)(-6) - (3)(-4) = -12 + 12 = 0$.
Thus,$|A| I = 0 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Next,find the adjoint of $A$ $(\text{adj } A)$:
The cofactors are $C_{11} = -6, C_{12} = 4, C_{21} = -3, C_{22} = 2$.
Therefore,$\text{adj } A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix}$.
Now,calculate $A(\text{adj } A)$:
$A(\text{adj } A) = \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} = \begin{bmatrix} -12+12 & -6+6 \\ 24-24 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Finally,calculate $(\text{adj } A) A$:
$(\text{adj } A) A = \begin{bmatrix} -6 & -3 \\ 4 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ -4 & -6 \end{bmatrix} = \begin{bmatrix} -12+12 & -18+18 \\ 8-8 & 12-12 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$.
Since $A(\text{adj } A) = (\text{adj } A) A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = |A| I$,the identity is verified.