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 \(\boxed{16}\)

Answer 2

To solve the problem, we analyze the set \(S\) of \(3 \times 3\) matrices of the form:

\[ A = \begin{pmatrix} 0 & 1 & c \\ 1 & a & d \\ 1 & b & e \end{pmatrix} \] where \(a, b, c, d, e \in \{0,1\}\), and the determinant \(|A| \in \{-1, 1\}\). We want to find how many such matrices have determinant \(\pm 1\).

 

Step 1: Total possible matrices
Each of the 5 variables \(a,b,c,d,e\) can be either 0 or 1.
Number of matrices: \[ 2^5 = 32 \]

Step 2: Expression for determinant \(|A|\)
Calculate \(|A|\) using cofactor expansion along the first row:

\[ |A| = 0 \times M_{11} - 1 \times M_{12} + c \times M_{13} \] where \(M_{ij}\) are minors.

 

Calculate minors:

\[ M_{12} = \begin{vmatrix} 1 & d \\ 1 & e \end{vmatrix} = 1 \cdot e - 1 \cdot d = e - d \] \[ M_{13} = \begin{vmatrix} 1 & a \\ 1 & b \end{vmatrix} = 1 \cdot b - 1 \cdot a = b - a \]

So determinant is:

\[ |A| = -1 \times (e - d) + c \times (b - a) = - (e - d) + c (b - a) = d - e + c(b - a) \] Since all variables are 0 or 1, each term can be 0 or 1, and differences are from \(\{-1,0,1\}\).

 

Step 3: Values of \(|A|\) for each possible combination
We want \(|A| = \pm 1\), so possible values are \(1\) or \(-1\).

Rewrite:

\[ |A| = d - e + c(b - a) \] All variables \(a,b,c,d,e\) are 0 or 1.

 

Step 4: Enumerate possibilities for \(c, b, a\):
Values of \(b - a\) can be:

| \(b\) | \(a\) | \(b - a\) | |-------|-------|---------| | 0 | 0 | 0 | | 0 | 1 | -1 | | 1 | 0 | 1 | | 1 | 1 | 0 |

Step 5: Calculate \(c(b - a)\):
Since \(c = 0\) or \(1\), \(c(b - a)\) is either 0 or \(b - a\).

Step 6: Now write \(|A| = d - e + c(b - a)\) = \(d - e + k\), where \(k = 0\) or \(b - a\).

Step 7: Possible values of \(d - e\):

| \(d\) | \(e\) | \(d - e\) | |-------|-------|---------| | 0 | 0 | 0 | | 0 | 1 | -1 | | 1 | 0 | 1 | | 1 | 1 | 0 |

Step 8: Now \(|A| = (d - e) + k\), and we want \(|A| = \pm 1\).

Case 1: \(k = 0\)
\(|A| = d - e\). So \(|A| = \pm 1\) if \(d - e = \pm 1\), i.e., if \(d=1, e=0\) or \(d=0, e=1\).

Case 2: \(k = 1\)
\(|A| = d - e + 1\). For \(|A| = 1\), we have:

\[ d - e + 1 = 1 \implies d - e = 0 \] So \(d = e\), either both 0 or both 1.

 

Case 3: \(k = -1\)
\(|A| = d - e - 1\). For \(|A| = -1\), we have:

\[ d - e - 1 = -1 \implies d - e = 0 \] Again, \(d = e\).

 

Step 9: Summarize:
- When \(k = 0\), \(|A| = \pm 1\) if \(d \neq e\) (2 cases).
- When \(k = \pm 1\), \(|A| = \pm 1\) if \(d = e\) (2 cases).
- When \(k\) takes values 0, 1, or -1 depending on \(c,b,a\).

Step 10: Count the number of \((a,b,c)\) tuples for each \(k\):
Recall from Step 4:

| \(b\) | \(a\) | \(b - a\) | |-------|-------|---------| | 0 | 0 | 0 | | 0 | 1 | -1 | | 1 | 0 | 1 | | 1 | 1 | 0 |

Number of \((a,b)\) pairs with:

- \(b - a = 0\): 2 pairs \((0,0)\) and \((1,1)\)
- \(b - a = 1\): 1 pair \((1,0)\)
- \(b - a = -1\): 1 pair \((0,1)\)

 

Since \(c\) can be 0 or 1:

- For \(k = 0\), when \(c=0\), \(k=0\) always.
- For \(k = b - a\), when \(c=1\), \(k = b - a\).

 

Thus: - \(k=0\) occurs when \(c=0\) (all 4 \((a,b)\) pairs) → 4 cases
- \(k=1\) occurs when \(c=1\) and \((b,a) = (1,0)\) → 1 case
- \(k=-1\) occurs when \(c=1\) and \((b,a) = (0,1)\) → 1 case
- \(k=0\) also occurs when \(c=1\) and \((b,a) = (0,0)\) or \((1,1)\) → 2 cases

Total \((a,b,c)\) combinations = \(4 \times 2 = 8\).

Step 11: For each \(k\), count valid \(d,e\) pairs:
- When \(k=0\), \(|A| = d - e\), so \(|A| = \pm 1\) if \(d \neq e\) → 2 valid \((d,e)\) pairs
- When \(k = \pm 1\), \(|A| = d - e \pm 1\), so \(|A| = \pm 1\) if \(d = e\) → 2 valid \((d,e)\) pairs (since \(d,e \in \{0,1\}\))

Step 12: Calculate total number of matrices in \(S\):
- For \(k=0\), number of \((a,b,c)\) cases = 6 (4 when \(c=0\) + 2 when \(c=1\) and \(b-a=0\))
Each has 2 valid \((d,e)\) pairs → \(6 \times 2 = 12\)
- For \(k=1\), number of \((a,b,c)\) cases = 1
Each has 2 valid \((d,e)\) pairs → \(1 \times 2 = 2\)
- For \(k=-1\), number of \((a,b,c)\) cases = 1
Each has 2 valid \((d,e)\) pairs → \(1 \times 2 = 2\)

Total valid matrices = \(12 + 2 + 2 = 16\).

Final Answer:
\[ \boxed{16} \]