3 and 4 .Determinants and Matrices
hard

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 $Z$ is set of all integers. Then the number of $A \in M$ such that $f(A)=15$ is equal to $.....$

A

$16$

B

$32$

C

$48$

D

$71$

(JEE MAIN-2021)

Solution

$|\mathrm{A}|=\mathrm{ad}-\mathrm{bc}=15$

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

Case $\mathrm{I} \mathrm{ad}=9 \,\& \,\mathrm{bc}=-6$

For ad possible pairs are $(3,3),(-3,-3)$

For bc possible pairs are $(3,-2),(-3,2),(-2,3),\left(2_{6}-3\right)$

So total matrix $=2 \times 4=8$

Case $II$ ad $=6 \,\&\, \mathrm{bc}=-9$

Similarly total matrix $=2 \times 4=8$

$\Rightarrow$ Total such matrices are $=16$

Standard 12
Mathematics

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