Question:
Let $M=\left\{A=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right): a, b, c, d \in\{\pm 3, \pm 2, \pm 1,0\}\right\}$
Define $f: M \rightarrow Z$, as $f(A)=\operatorname{det}(A)$, for all $A \in M$, where $\mathbf{Z}$ is set of all integers. Then the number of $A \in M$ such that $f(A)=15$ is equal to_________.
Solution:
$|A|=a d-b c=15$
where $a, b, c, d \in\{\pm 3, \pm 2, \pm 1,0\}$
Case I ad $=9 \& \mathrm{bc}=-6$
For ad possible pairs are $(3,3),(-3,-3)$
For bc possible pairs are $(3,-2),(-3,2),(-2,3),(2,-3)$
So total matrix $=2 \times 4=8$
Case II $\mathrm{ad}=6 \quad \& \quad \mathrm{bc}=-9$
Similarly total matrix $=2 \times 4=8$
$\Rightarrow$ Total such matrices are $=16$