Question:

Given $A = \begin{pmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{pmatrix}$ where $a, b, c, d, e \in \{0, 1\}$.<br>The determinant $|A| = 0(ae - bd) - 1(e - d) + c(b - a) = d - e + c(b - a)$.<br>We are given $|A| \in \{-1, 1\}$.<br>Case 1: $c = 0$<br>$|A| = d - e$<br>If $|A| = 1$, then $d = 1, e = 0$. $a, b$ can be any of $\{0, 1\}$, so $2 \times 2 = 4$ matrices.<br>If $|A| = -1$, then $d = 0, e = 1$. $a, b$ can be any of $\{0, 1\}$, so $2 \times 2 = 4$ matrices.<br>Total for $c = 0$ is $4 + 4 = 8$.<br>Case 2: $c = 1$<br>$|A| = d - e + b - a$<br>Let $x = d - e$ and $y = b - a$. Note that $x, y \in \{-1, 0, 1\}$.<br>We want $x + y \in \{-1, 1\}$.<br>Subcase 2.1: $x + y = 1$<br>Possible $(x, y)$ pairs are $(1, 0)$ and $(0, 1)$.<br>- For $(1, 0)$: $(d, e) = (1, 0)$ (1 way) and $(b, a) \in \{(0, 0), (1, 1)\}$ (2 ways). Total $1 \times 2 = 2$.<br>- For $(0, 1)$: $(d, e) \in \{(0, 0), (1, 1)\}$ (2 ways) and $(b, a) = (1, 0)$ (1 way). Total $2 \times 1 = 2$.<br>Total for $x + y = 1$ is $2 + 2 = 4$.<br>Subcase 2.2: $x + y = -1$<br>Possible $(x, y)$ pairs are $(-1, 0)$ and $(0, -1)$.<br>- For $(-1, 0)$: $(d, e) = (0, 1)$ (1 way) and $(b, a) \in \{(0, 0), (1, 1)\}$ (2 ways). Total $1 \times 2 = 2$.<br>- For $(0, -1)$: $(d, e) \in \{(0, 0), (1, 1)\}$ (2 ways) and $(b, a) = (0, 1)$ (1 way). Total $2 \times 1 = 2$.<br>Total for $x + y = -1$ is $2 + 2 = 4$.<br>Total for $c = 1$ is $4 + 4 = 8$.<br>Total number of elements in $S = 8 + 8 = 16$.