Question

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

Answer 1

Correct Answer: 16

Let \(A = \begin{pmatrix} 0 & 1 & c \\ 1 & a & d\\ 1 & b & e \end{pmatrix}\), where \(a, b, c, d, e \in \{0, 1\}\). 
We need to find the number of such matrices \(A\) with \(|A| \in \{-1, 1\}\). 
The determinant of \(A\) is: \(|A| = 0(ae - bd) - 1(e - d) + c(b - a) = -(e - d) + c(b - a) = d - e + c(b - a)\). 
Since \(a, b, c, d, e \in \{0, 1\}\), the possible values of \(b - a\) are \(-1, 0, 1\). Also, \(d - e\) can be \(-1, 0, 1\). 
We want \(|A| \in \{-1, 1\}\), so \(d - e + c(b - a) = \pm 1\).

Case 1: \(c = 0\). Then \(d - e = \pm 1\). 
If \(d - e = 1\), then \(d = 1, e = 0\). If \(d - e = -1\), then \(d = 0, e = 1\). 
In this case, \(c = 0\), \(d - e = \pm 1\). 
\(a\) and \(b\) can take any value in \(\{0, 1\}\), so we have 2 choices for \(a\) and 2 choices for \(b\).
Therefore, we have \(1 \cdot 2 \cdot 2 \cdot 2 = 8\) matrices.

Case 2: \(c = 1\). 
Then \(d - e + b - a = \pm 1\). \(d - e + b - a = 1 \Rightarrow d + b = 1 + e + a\). 
\(d - e + b - a = -1 \Rightarrow d + b = e + a - 1\). \(d + b = 1 + e + a\): 
- \(d + b = 0\): \(d = b = 0\). 
Then \(e + a = -1\), which is impossible. 
- \(d + b = 1\): \((d = 1, b = 0)\) or \((d = 0, b = 1)\). 
If \(d = 1, b = 0\), then \(e + a = 0\), so \(e = 0, a = 0\). 
(1, 0, 0, 0) If \(d = 0, b = 1\), then \(e + a = 0\), so \(e = 0, a = 0\). 
(0, 1, 0, 0) - \(d + b = 2\): \(d = b = 1\). 
Then \(e + a = 1\), so \((e, a) = (1, 0), (0, 1)\). (1, 1, 1, 0), (1, 1, 0, 1)

\(d + b = e + a - 1\): Since \(e+a\) must be at least \(0\), we must have \(e+a = 0\), \(1\), or \(2\). 
Thus, \(e+a-1 = -1, 0, 1\). 
- \(d+b = -1\): impossible since \(d, b \ge 0\). - \(d+b = 0\): \(d = b = 0\). 
Then \(e + a = 1\), so \((e, a) = (1, 0), (0, 1)\). (0, 0, 1, 0), (0, 0, 0, 1) - \(d+b = 1\): 
\((d = 1, b = 0)\) or \((d = 0, b = 1)\). 
Then \(e + a = 2\), so \(e = 1, a = 1\). (1, 0, 1, 1), (0, 1, 1, 1) - \(d+b = 2\): 
\(d = b = 1\). 
Then \(e + a = 3\), impossible.

In total, we have the following values for (d, b, e, a): 
(1, 0, 0, 0), (0, 1, 0, 0), (1, 1, 1, 0), (1, 1, 0, 1) (0, 0, 1, 0), (0, 0, 0, 1), (1, 0, 1, 1), (0, 1, 1, 1) 
So we have 8 cases. If \(c = 1\), there are 8 possibilities for \((a,b,d,e)\). 
Hence the number of elements in S is \(8+8 = 16\).

Final Answer: The final answer is \(\boxed{16}\)