If A is a non-diagonal involutory matrix, then

Options

A. `A-I=O`
B. `A+I=O`
C. `A-I` is nonzero singular
D. none of these

Explanation

To solve the problem, we need to analyze the properties of the non-diagonal involutory matrix \( A \). An involutory matrix is defined as a matrix that satisfies the equation \( A^2 = I \), where \( I \) is the identity matrix.

### Step-by-step Solution:

1. **Definition of Involutory Matrix**:
Since \( A \) is an involutory matrix, we have:
\[ A^2 = I \]

2. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ A^2 - I = 0 \]
This can be factored as:
\[ (A - I)(A + I) = 0 \]

3. **Taking Determinants**:
Taking the determinant of both sides, we get:
\[ \text{det}((A - I)(A + I)) = \text{det}(0) = 0 \]
Using the property of determinants, we can write:
\[ \text{det}(A - I) \cdot \text{det}(A + I) = 0 \]

4. **Analyzing the Factors**:
This implies that at least one of the following must be true:
\[ \text{det}(A - I) = 0 \quad \text{or} \quad \text{det}(A + I) = 0 \]

5. **Considering Non-Diagonal Matrix**:
We know that \( A \) is a non-diagonal matrix, meaning not all off-diagonal elements are zero. We will analyze both cases:

- **Case 1**: If \( \text{det}(A - I) = 0 \), then \( A - I \) is a singular matrix. This means that \( A \) has at least one eigenvalue equal to 1.

- **Case 2**: If \( \text{det}(A + I) = 0 \), then \( A + I \) is also a singular matrix. This means that \( A \) has at least one eigenvalue equal to -1.

6. **Conclusion**:
Since \( A \) is a non-diagonal involutory matrix, it cannot be the case that both \( A - I \) and \( A + I \) are non-singular. Thus, we conclude that at least one of the determinants must be zero, leading us to the conclusion that \( A - I \) is a non-zero singular matrix.

### Final Result:
Thus, we can conclude that if \( A \) is a non-diagonal involutory matrix, then \( A - I \) is a non-zero singular matrix.

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