If A `= (a_(ij))_(3xx)` is a matrix satrisfying the equation `x^(3) -3x+1 =0` then

Options

A. A is a unit matrix
B. A is singular matrix
C. A is non -singular matrix
D. none of these

Answer

To solve the problem, we need to analyze the matrix \( A \) which satisfies the equation \( A^3 - 3A + I = 0 \), where \( I \) is the identity matrix of order 3. We will derive the determinant of \( A \) from this equation.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ A^3 - 3A + I = 0 \]
Rearranging gives:
\[ A^3 - 3A = -I \]

2. **Factor out \( A \):**
We can rewrite the equation as:
\[ A^3 - 3A I = -I \]
This implies:
\[ A^3 - 3A I + I = 0 \]

3. **Take determinants on both sides:**
Taking the determinant of both sides, we have:
\[ \det(A^3 - 3A I) = \det(-I) \]

4. **Use the property of determinants:**
The determinant of a product of matrices is the product of their determinants:
\[ \det(A^3 - 3A I) = \det(A^3) \cdot \det(I - 3A^{-1}) \]
Since \( \det(-I) = (-1)^3 \cdot \det(I) = -1 \).

5. **Calculate \( \det(A^3) \):**
Using the property of determinants, we know:
\[ \det(A^3) = (\det(A))^3 \]

6. **Substituting back:**
We can substitute this back into our determinant equation:
\[ (\det(A))^3 \cdot \det(I - 3A^{-1}) = -1 \]

7. **Evaluate \( \det(I - 3A^{-1}) \):**
Since \( A \) is a 3x3 matrix, we know that \( \det(I) = 1 \). Therefore:
\[ \det(I - 3A^{-1}) = 1 - 3 \det(A^{-1}) = 1 - \frac{3}{\det(A)} \]

8. **Final equation:**
Combining everything, we have:
\[ (\det(A))^3 \left(1 - \frac{3}{\det(A)}\right) = -1 \]

9. **Conclusion:**
From this equation, we can conclude that \( \det(A) \neq 0 \), which indicates that the matrix \( A \) is non-singular.