If A = `[ (alpha,2),(2 , alpha)]` and det `(A^(3))` = 125 then `alpha` is equal to

Options

A. `pm 1`
B. `pm 2`
C. `pm 3`
D. `pm 5`

Answer

To solve the problem, we need to find the value of \( \alpha \) given the matrix \( A = \begin{pmatrix} \alpha & 2 \\ 2 & \alpha \end{pmatrix} \) and the condition that \( \text{det}(A^3) = 125 \).

### Step-by-Step Solution:

1. **Understanding the Determinant of \( A^3 \)**:
We know that the determinant of a matrix raised to a power can be expressed as:
\[ \text{det}(A^n) = (\text{det}(A))^n \]
Therefore, we have:
\[ \text{det}(A^3) = (\text{det}(A))^3 \]
Given that \( \text{det}(A^3) = 125 \), we can write:
\[ (\text{det}(A))^3 = 125 \]

2. **Finding the Determinant of \( A \)**:
The determinant of matrix \( A \) is calculated as follows:
\[ \text{det}(A) = \alpha \cdot \alpha - 2 \cdot 2 = \alpha^2 - 4 \]

3. **Setting Up the Equation**:
From step 1, we have:
\[ (\text{det}(A))^3 = 125 \]
Substituting the expression for \( \text{det}(A) \):
\[ (\alpha^2 - 4)^3 = 125 \]

4. **Taking the Cube Root**:
Taking the cube root of both sides gives:
\[ \alpha^2 - 4 = 5 \]

5. **Solving for \( \alpha^2 \)**:
Adding 4 to both sides:
\[ \alpha^2 = 5 + 4 = 9 \]

6. **Finding \( \alpha \)**:
Taking the square root of both sides:
\[ \alpha = \pm 3 \]

### Final Answer:
Thus, the values of \( \alpha \) are \( 3 \) and \( -3 \).