JEE Advanced 2024 Paper 1 · Q10 · Determinants

Question

Let $S = \left\{ A = \begin{pmatrix} 0 & 1 & c \ 1 & a & d \ 1 & b & e \end{pmatrix} : a, b, c, d, e \in \{0, 1\}\ 	ext{and}\ |A| \in \{-1, 1\} ight\}$, where $|A|$ denotes the determinant of $A$. Then the number of elements in $S$ is ______.

SolveKar AI · Accepted Answer

Expand $|A|$ along row 1: $|A| = 0\cdot M_{11} - 1\cdot(e-d) + c\cdot(b-a) = -(e-d)+c(b-a) = (d-e)+c(b-a)$. We need $|A|\in\{-1,1\}$ with $a,b,c,d,e\in\{0,1\}$. Case $c=0$: $|A|=d-e\in\{-1,0,1\}$. Need $\pm 1$: $(d,e)\in\{(1,0),(0,1)\}$, with $a,b$ free $\Rightarrow 2\cdot 4=8$. Case $c=1$: $|A|=(d-e)+(b-a)$. Each term $\in\{-1,0,1\}$. We want sum $\pm 1$. Enumerate: sum$=1$: $(1,0)+(0,0)$, $(0,0)+(1,0)$, $(0,-1)+... $. Cleanest: count pairs $(d-e,b-a)$ summing to $\pm 1$. $(d-e)$ over $(d,e)$: values $\{-1,0,0,1\}$ (counts 1,2,1). Same for $(b-a)$. Sum $=1$: $(1,0)$ has $1\cdot 2=2$; $(0,1)$ has $2\cdot 1=2$; total $4$. Symmetric for $-1$: $4$. So Case $c=1$ gives $8$. Total $|S|=8+8=16$.

JEE Advanced 2024 Paper 1 · Q10 · Determinants

Solution