Answer
$16$
Work Step by Step
The general form of a matrix of order $ 2 \times 2$ is:
$det \ A=\begin{bmatrix} p & q \\ r & s\end{bmatrix}=ps-qr$
Since Column-3 has the most zeroes, we choose Column-1.
Now, $det \ A =1 \begin{bmatrix} 2 & 3 \\ -5 & -1 \end{bmatrix} -0+3 \begin{bmatrix} -1 & -2 \\ 2 & 3 \end{bmatrix} =-2+15-0+3(-3+4)=16$
