If $c_{ij}$ denotes the cofactor of element $a_{ij}$ of the matrix $A = \begin{bmatrix} 1 & 2 & -1 \\ 0 & -3 & 2 \\ 4 & 2 & 3 \end{bmatrix}$ then the value of $c_{21} \cdot c_{33}$ is

The cofactor $c_{ij}$ of an element $a_{ij}$ in a matrix is given by:

$c_{ij} = (-1)^{i+j} \cdot M_{ij}$

where $M_{ij}$ is the minor — the determinant of the smaller matrix obtained after deleting row $i$ and column $j$.

Given matrix:

$A = \begin{bmatrix} 1 & 2 & -1 \\ 0 & -3 & 2 \\ 4 & 2 & 3 \end{bmatrix}$

To find $c_{21}$, delete Row 2 and Column 1:

$M_{21} = \begin{vmatrix} 2 & -1 \\ 2 & 3 \end{vmatrix}$

$M_{21} = (2)(3) - (-1)(2) = 6 + 2 = 8$

$c_{21} = (-1)^{2+1} \cdot 8 = (-1)^3 \cdot 8 = -8$

To find $c_{33}$, delete Row 3 and Column 3:

$M_{33} = \begin{vmatrix} 1 & 2 \\ 0 & -3 \end{vmatrix}$

$M_{33} = (1)(-3) - (2)(0) = -3 - 0 = -3$

$c_{33} = (-1)^{3+3} \cdot (-3) = (-1)^6 \cdot (-3) = -3$

$c_{21} \cdot c_{33} = (-8) \times (-3) = \boxed{24}$