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3 and 4 .Determinants and Matrices
medium
Let $S=\left\{A=\left(\begin{array}{lll}0 & 1 & c \\ 1 & a & d \\ 1 & b & e\end{array}\right): a, b, c, d, e \in\{0,1\}\right.$ and $\left.|A| \in\{-1,1\}\right\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is. . . . .
A
$10$
B
$15$
C
$16$
D
$20$
(IIT-2024)
Solution
$|A|=0(a e-b d)-1(e-d)+c(b-a)$
$=c(b-a)+(d-e)$
$|A| \in\{-1,1\}$ and $a, b, c, d, e \in\{0,1\}$
Case-$I$
$c =0 \quad d =1, e =0, a , b \in(0,1)$
$d =0, e =1$
$a b b d e$
$2 \downarrow$
$2.21 \quad 2 \rightarrow 8 \text { cases }$
Case-$II$
$c =1 \quad b =1, a =0, \quad d =0, e =0, d =1, e =1$
$b =0, a =1, \quad d =0, e =0, d =1, e =1$
$b =0, a =0, \quad d =1, e =0$
$d =1, a =1, \quad d =1, e =0$
$d =0, e =1$
$\rightarrow 8$ cases
$\rightarrow 8$ cases
$\Rightarrow$ Total $16$ cases
Standard 12
Mathematics