Answer
$28$
Work Step by Step
Formula to determine the determinant, $D$ of a $3 \times 3$ matrix is:
$D=\begin{vmatrix}a&b&c\\d&e&f\\g&h&i\end{vmatrix}=a \begin{vmatrix}e&f\\h&i\end{vmatrix}-b \begin{vmatrix}d&f\\g&i\end{vmatrix}+c \begin{vmatrix}d&e\\g&h\end{vmatrix}$
Thus, as per question, we have
$D=\begin{vmatrix}4&0&0\\3&-1&4\\2&-3&5\end{vmatrix}=3(-5+12)+0+0=28$
