To solve the problem, we need to determine the number of singular matrices in the set \( A \), where \( A \) is defined as:
\[
A = \begin{pmatrix}
a_{11} & a_{12} \\
a_{21} & a_{22}
\end{pmatrix}
\]
with the conditions that \( a_{ij} \in \{0, 1, 2\} \) and \( a_{11} = a_{22} \).
### Step 1: Determine the condition for singularity
A matrix is singular if its determinant is zero. The determinant of the matrix \( A \) is given by:
\[
\text{det}(A) = a_{11} a_{22} - a_{12} a_{21}
\]
Given that \( a_{11} = a_{22} \), we can rewrite the determinant as:
\[
\text{det}(A) = a_{11}^2 - a_{12} a_{21}
\]
### Step 2: Set the determinant to zero
For the matrix to be singular, we need:
\[
a_{11}^2 - a_{12} a_{21} = 0
\]
This simplifies to:
\[
a_{11}^2 = a_{12} a_{21}
\]
### Step 3: Analyze the possible values of \( a_{11} \)
Since \( a_{ij} \) can take values from the set \( \{0, 1, 2\} \), we will analyze each case for \( a_{11} \).
#### Case 1: \( a_{11} = 0 \)
If \( a_{11} = 0 \), then:
\[
0 = a_{12} a_{21}
\]
This means either \( a_{12} = 0 \) or \( a_{21} = 0 \) (or both). The combinations are:
1. \( a_{12} = 0, a_{21} = 0 \) (1 way)
2. \( a_{12} = 0, a_{21} = 1 \) (1 way)
3. \( a_{12} = 0, a_{21} = 2 \) (1 way)
4. \( a_{12} = 1, a_{21} = 0 \) (1 way)
5. \( a_{12} = 2, a_{21} = 0 \) (1 way)
Total for this case: 5 ways.
#### Case 2: \( a_{11} = 1 \)
If \( a_{11} = 1 \), then:
\[
1 = a_{12} a_{21}
\]
The valid combinations are:
1. \( a_{12} = 1, a_{21} = 1 \) (1 way)
Total for this case: 1 way.
#### Case 3: \( a_{11} = 2 \)
If \( a_{11} = 2 \), then:
\[
4 = a_{12} a_{21}
\]
The valid combinations are:
1. \( a_{12} = 2, a_{21} = 2 \) (1 way)
Total for this case: 1 way.
### Step 4: Sum the total number of singular matrices
Now, we add the total number of ways from all cases:
- Case 1: 5 ways
- Case 2: 1 way
- Case 3: 1 way
Total singular matrices = \( 5 + 1 + 1 = 7 \).
### Final Answer
The number of singular matrices in the set \( A \) is \( \boxed{7} \).
