- Home
- Standard 12
- Mathematics
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