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

Options

A. `+-1`
B. `+-2`
C. `+-3`
D. ` +-5`

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. The matrix \( A \) is defined as:

\[ A = \begin{pmatrix} \alpha & 2 \\ 2 & \alpha \end{pmatrix} \]

### Step 1: Find the determinant of matrix \( A \)

The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by the formula:

\[ \text{det}(A) = ad - bc \]

For our matrix \( A \):

\[ \text{det}(A) = \alpha \cdot \alpha - 2 \cdot 2 = \alpha^2 - 4 \]

### Step 2: Relate the determinant of \( A^3 \) to \( |A| \)

We know that the determinant of \( A^n \) is given by:

\[ |A^n| = |A|^n \]

In our case, we have \( n = 3 \):

\[ |A^3| = |A|^3 \]

Given that \( |A^3| = 125 \), we can write:

\[ |A|^3 = 125 \]

### Step 3: Solve for \( |A| \)

Taking the cube root of both sides, we find:

\[ |A| = \sqrt[3]{125} = 5 \]

### Step 4: Set up the equation

Now we can set up the equation using our expression for \( |A| \):

\[ \alpha^2 - 4 = 5 \]

### Step 5: Solve for \( \alpha \)

Now, we solve the equation:

\[ \alpha^2 - 4 - 5 = 0 \]
\[ \alpha^2 - 9 = 0 \]
\[ \alpha^2 = 9 \]

Taking the square root of both sides, we get:

\[ \alpha = \pm 3 \]

### Conclusion

Thus, the values of \( \alpha \) are \( 3 \) and \( -3 \).