Answer
See below
Work Step by Step
We have $sum A=12+3+5+4=24$
We obtain reduced row echelon matrix form:
$\begin{bmatrix}
12 & 11 & 9 & -7\\
0 & \frac{7}{6} & - \frac{13}{2} & \frac{43}{6}\\
0 & 0 & -9 & 17\\
0 & 0 & 0 & \frac{607}{126}
\end{bmatrix}$
Product of the given matrix is: $\det (A)=-607$
