Let M = A = a b c d : a , b , c , d ∈ ± 3 , ± 2 , ± 1 , 0 . Define f : M → Z , as…

.$\left|A\right|=ad-bc=15$

where $a,b,c,d\in \left\{\pm 3,\pm 2,\pm 1,0\right\}$

Case $\mathrm{I}:$ $ad=9 \& bc=-6$

For $ad$ possible pairs are $\left(3,3\right),\left(-3,-3\right).$ 

For $bc$ possible pairs are $\left(3,-2\right),\left(-3,2\right),\left(-2,3\right),\left(2,-3\right)$ 

So, total number of matrices in case $\mathrm{I}$$=2\times 4=8$

Case $\mathrm{II}:$ $ad=6 \& bc=-9$

Similarly, total number of matrices in case $\mathrm{II}$ $=2\times 4=8$

Hence, total number of matrices are $16$.