Number of Binary Matrices with Determinant Absolute Value One

To find the number of elements in the set $S$, we first evaluate the determinant of the given $3 \times 3$ matrix $A$:

$A = \begin{pmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{pmatrix}$

Expanding the determinant along the first row:

$|A| = 0 \cdot (ae - bd) - 1 \cdot (e - d) + c \cdot (b - a)$ $|A| = d - e + c(b - a)$

We are given that $a, b, c, d, e \in \{0, 1\}$ and $|A| \in \{-1, 1\}$. We analyze the possible cases based on the value of $c$:

---

Case 1: $c = 0$

When $c = 0$, the expression simplifies to: $|A| = d - e$

For $|A| \in \{-1, 1\}$, we must have:

1. $d - e = 1 \implies (d, e) = (1, 0)$
2. $d - e = -1 \implies (d, e) = (0, 1)$

Thus, there are $2$ valid pairs for $(d, e)$. Since $a$ and $b$ can independently take any value in $\{0, 1\}$, there are $2 \times 2 = 4$ choices for $(a, b)$.

$\text{Number of elements for } c = 0 \text{ is } 2 \times 4 = 8$

---

Case 2: $c = 1$

When $c = 1$, the expression becomes: $|A| = (d - e) + (b - a)$

Let $x = d - e$ and $y = b - a$. Since $a, b, d, e \in \{0, 1\}$, the possible values of $x$ and $y$ are $\{-1, 0, 1\}$:

• $x = 1 \implies (d, e) = (1, 0)$ (1 pair)
• $x = 0 \implies (d, e) \in \{(0, 0), (1, 1)\}$ (2 pairs)
• $x = -1 \implies (d, e) = (0, 1)$ (1 pair)

Similarly for $y = b - a$:

• $y = 1 \implies (b, a) = (1, 0)$ (1 pair)
• $y = 0 \implies (b, a) \in \{(0, 0), (1, 1)\}$ (2 pairs)
• $y = -1 \implies (b, a) = (0, 1)$ (1 pair)

We require $|A| = x + y \in \{-1, 1\}$:

1. For $x + y = 1$:

   • $x = 1, y = 0 \implies 1 \times 2 = 2$ combinations of $(a, b, d, e)$
   • $x = 0, y = 1 \implies 2 \times 1 = 2$ combinations of $(a, b, d, e)$
   • Subtotal = $2 + 2 = 4$

2. For $x + y = -1$:

   • $x = -1, y = 0 \implies 1 \times 2 = 2$ combinations of $(a, b, d, e)$
   • $x = 0, y = -1 \implies 2 \times 1 = 2$ combinations of $(a, b, d, e)$
   • Subtotal = $2 + 2 = 4$

$\text{Number of elements for } c = 1 \text{ is } 4 + 4 = 8$

---

Total Count

Summing the valid combinations from both cases: $\text{Total elements in } S = 8 + 8 = 16$