If `A=[{:(alpha,2),(2, alpha):}]` and `|A^(3)|=125` then the value of `alpha `is

Options

A. `pm1`
B. `pm2`
C. `pm3`
D. `pm5`

Answer

To solve the problem, we need to find the value of \( \alpha \) given that the determinant of the matrix \( A \) raised to the power of 3 is equal to 125.

Given:
\[ A = \begin{pmatrix} \alpha & 2 \\ 2 & \alpha \end{pmatrix} \]
and
\[ |A^3| = 125. \]

### Step 1: Use the property of determinants
We know that the determinant of a matrix raised to a power can be expressed as:
\[ |A^n| = |A|^n. \]
Thus, we can write:
\[ |A^3| = |A|^3. \]
Given \( |A^3| = 125 \), we have:
\[ |A|^3 = 125. \]

### Step 2: Find the determinant of matrix \( A \)
Next, we need to find the determinant of matrix \( A \):
\[ |A| = \begin{vmatrix} \alpha & 2 \\ 2 & \alpha \end{vmatrix} = \alpha \cdot \alpha - 2 \cdot 2 = \alpha^2 - 4. \]

### Step 3: Set up the equation
Now substituting \( |A| \) into the equation we derived:
\[ (\alpha^2 - 4)^3 = 125. \]

### Step 4: Take the cube root
Taking the cube root of both sides gives:
\[ \alpha^2 - 4 = 5. \]

### Step 5: Solve for \( \alpha \)
Now, we can solve for \( \alpha \):
\[ \alpha^2 - 4 = 5 \implies \alpha^2 = 5 + 4 \implies \alpha^2 = 9. \]
Taking the square root of both sides, we find:
\[ \alpha = \pm 3. \]

### Final Answer
Thus, the values of \( \alpha \) are:
\[ \alpha = 3 \quad \text{or} \quad \alpha = -3. \]

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