JEE Advanced · Mathematics · 19. Determinants
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 8
- B 12
- C 16
- D 20
Answer & Solution
Correct Answer
(C) 16
Step-by-step Solution
Detailed explanation
\(|A|=-(e-d)+c(b-a)= \pm 1\)
Case (i) : \(c=0 \Rightarrow(e, d)=(1,0),(0,1) \rightarrow 2\) ways
b and a can be each 2 ways
\(\Rightarrow\) Total \(=8\) ways
Case (ii) : \(c \neq 0 \Rightarrow c=1\)
\(\Rightarrow d-e+b-a= \pm 1\)
\(\left.\begin{array}{llll}1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right\} \rightarrow 4 \times 2=8\) ways
Total \(=16\) ways
Case (i) : \(c=0 \Rightarrow(e, d)=(1,0),(0,1) \rightarrow 2\) ways
b and a can be each 2 ways
\(\Rightarrow\) Total \(=8\) ways
Case (ii) : \(c \neq 0 \Rightarrow c=1\)
\(\Rightarrow d-e+b-a= \pm 1\)
\(\left.\begin{array}{llll}1 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right\} \rightarrow 4 \times 2=8\) ways
Total \(=16\) ways
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