Let `A` be a square matrix of `2 x 2` satisfyinga. `a_(ij)=1` or `-1` and `a_11*a_21+a_12*a_22=0` then the no. of matrix

Answer

To solve the problem, we need to find the number of \(2 \times 2\) matrices \(A\) such that each element \(a_{ij}\) is either \(1\) or \(-1\) and the condition \(a_{11} \cdot a_{21} + a_{12} \cdot a_{22} = 0\) holds.

Let's denote the matrix \(A\) as follows:

\[ A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \]

### Step 1: Analyze the condition
The condition we need to satisfy is:

\[ a_{11} \cdot a_{21} + a_{12} \cdot a_{22} = 0 \]

This implies that the two products \(a_{11} \cdot a_{21}\) and \(a_{12} \cdot a_{22}\) must be equal in magnitude but opposite in sign.

### Step 2: Possible values for elements
Since each element \(a_{ij}\) can either be \(1\) or \(-1\), we will consider the combinations of these values.

### Step 3: Case analysis
We can analyze the cases based on the values of \(a_{11}\) and \(a_{12}\):

1. **Case 1**: \(a_{11} = 1\)
- If \(a_{21} = 1\), then \(a_{12} \cdot a_{22} = -1\). This can happen if:
- \(a_{12} = 1\) and \(a_{22} = -1\)
- \(a_{12} = -1\) and \(a_{22} = 1\)
- If \(a_{21} = -1\), then \(a_{12} \cdot a_{22} = 1\). This can happen if:
- \(a_{12} = 1\) and \(a_{22} = 1\)
- \(a_{12} = -1\) and \(a_{22} = -1\)

2. **Case 2**: \(a_{11} = -1\)
- If \(a_{21} = 1\), then \(a_{12} \cdot a_{22} = 1\). This can happen if:
- \(a_{12} = 1\) and \(a_{22} = 1\)
- \(a_{12} = -1\) and \(a_{22} = -1\)
- If \(a_{21} = -1\), then \(a_{12} \cdot a_{22} = -1\). This can happen if:
- \(a_{12} = 1\) and \(a_{22} = -1\)
- \(a_{12} = -1\) and \(a_{22} = 1\)

### Step 4: Count the valid combinations
From the above cases, we can summarize the valid combinations:

- For \(a_{11} = 1\):
- \(a_{21} = 1\): (1, -1), (-1, 1) → 2 combinations
- \(a_{21} = -1\): (1, 1), (-1, -1) → 2 combinations
- For \(a_{11} = -1\):
- \(a_{21} = 1\): (1, 1), (-1, -1) → 2 combinations
- \(a_{21} = -1\): (1, -1), (-1, 1) → 2 combinations

Thus, we have a total of \(2 + 2 + 2 + 2 = 8\) valid matrices.

### Final Answer
The total number of \(2 \times 2\) matrices \(A\) that satisfy the given conditions is **8**.

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