Matrices and Determinants
PYP_JEE_ADV_2024_P1
Grade None

Question:

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

Step-by-Step Solution

Key Concept: Expand determinant along first row; then enumerate by c=0 and c=1 separately
$|A|=0(ae-bd)-1(e-d)+c(b-a)=d-e+c(b-a)$. We need $d-e+c(b-a)\in\{-1,1\}$. Total parameters: $a,b,c,d,e\in\{0,1\}$, so 32 cases. $c=0$: need $d-e=\pm1$. Exactly 2 of the 4 $(d,e)$ pairs work: $(1,0)$ and $(0,1)$. With $a,b$ free (4 choices): $4\times2=8$. $c=1$: need $d-e+(b-a)=\pm1$. - $(a,b)=(0,0)$: $b-a=0$, need $d-e=\pm1$: 2 cases. - $(a,b)=(1,0)$: $b-a=-1$, need $d-e=2$ (impossible) or $d-e=0$: $d=e$, 2 cases. - $(a,b)=(0,1)$: $b-a=1$, need $d-e=0$: 2 cases. - $(a,b)=(1,1)$: $b-a=0$, need $d-e=\pm1$: 2 cases. Subtotal: $2+2+2+2=8$. Total: $8+8=16$.
Correct Answer: 16

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