JEE Advanced 2024 Paper 1, Mathematics Section 3 Q3: Algebra Of Determinants

Expanding along the first row, $|A|=0\cdot M_{11}-1\cdot (e-d)+c\cdot (b-a)=-(e-d)+c(b-a)$, which must be $\pm 1$ with all entries in $\{0,1\}$.

Case 1: $c=0$. Then $|A|=d-e=\pm 1$. With $d,e\in \{0,1\}$, the pairs $(d,e)$ are $(1,0)$ or $(0,1)$, giving 2 choices. The entries $a,b$ are free in $\{0,1\}$, giving $2\times 2=4$ choices. Total: $2\times 4=8$.

Case 2: $c=1$. Then $|A|=(d-e)+(b-a)=\pm 1$. We need integer pairs $(d-e,b-a)$ from $\{-1,0,1\}$ summing to $\pm 1$. The valid combinations are:

$(1,0),(0,1),(-1,0),(0,-1)$. Each value $0$ for a difference can come from $(0,0)$ or $(1,1)$ (2 ways), while $\pm 1$ has 1 way each.

Counting: $(1,0)\to 1\times 2=2$; $(0,1)\to 2\times 1=2$; $(-1,0)\to 1\times 2=2$; $(0,-1)\to 2\times 1=2$. Total: $8$.

$$\therefore |S|=8+8=16$$