Question 10

The determinant $|A| = 0(ae - bd) - 1(e - d) + c(b - a) = d - e + c(b - a)$. We need $|A| = 1$ or $|A| = -1$ with $a, b, c, d, e \in \{0, 1\}$. Case 1: $c = 0$. $|A| = d - e$. For $|A| = 1$, $(d,e) = (1,0)$. $a, b$ can be any of $\{0,1\}$, so $2 imes 2 = 4$ ways. For $|A| = -1$, $(d,e) = (0,1)$. $a, b$ can be any of $\{0,1\}$, so $2 imes 2 = 4$ ways. Total for $c=0$ is $4 + 4 = 8$.

Case 2: $c = 1$. $|A| = d - e + b - a$. Let $X = d + b$ and $Y = e + a$. Possible values for $X, Y$ are $\{0, 1, 2\}$. $|A| = X - Y$. For $|A| = 1$, $(X,Y) \in \{(1,0), (2,1)\}$.

• $X=1, Y=0 \implies (d,b) \in \{(1,0), (0,1)\}$ and $(e,a)=(0,0)$. Total $2 \times 1 = 2$ ways.
• $X=2, Y=1 \implies (d,b)=(1,1)$ and $(e,a) \in \{(1,0), (0,1)\}$. Total $1 \times 2 = 2$ ways. Subtotal for $|A|=1$ is 4. For $|A| = -1$, $(X,Y) \in \{(0,1), (1,2)\}$.
• $X=0, Y=1 \implies (d,b)=(0,0)$ and $(e,a) \in \{(1,0), (0,1)\}$. Total $1 \times 2 = 2$ ways.
• $X=1, Y=2 \implies (d,b) \in \{(1,0), (0,1)\}$ and $(e,a)=(1,1)$. Total $2 \times 1 = 2$ ways. Subtotal for $|A|=-1$ is 4. Total for $c=1$ is $4 + 4 = 8$.

Overall total elements in $S = 8 + 8 = 16$.