Question

If \( A = \begin{bmatrix} k & 2 \\ 2 & k \end{bmatrix} \) and \( |A^3| = 125 \), then the value of \( k \) is:

• A. \( \pm 3 \)
• B. -5
• C. -4
• D. \( \pm 2 \)

Hint:

To find the determinant of a 2x2 matrix \( \begin{bmatrix} a & b
c & d \end{bmatrix} \), use the formula \( |A| = ad - bc \).

Answer 1

The Correct Option is A

If \( A = \begin{bmatrix} k & 2 \\ 2 & k \end{bmatrix} \) and \( |A^3| = 125 \), then the value of \( k \) is:

Step 1: Calculate the determinant of \( A \)
The determinant of a 2x2 matrix \( A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \) is given by:
\[ |A| = ad - bc \] For the matrix \( A = \begin{bmatrix} k & 2 \\ 2 & k \end{bmatrix} \), we can calculate the determinant \( |A| \): \[ |A| = k \times k - 2 \times 2 = k^2 - 4 \]

Step 2: Use the property of determinants
The property of determinants states that for any square matrix \( A \), the determinant of \( A^n \) is equal to \( |A|^n \). So: \[ |A^3| = |A|^3 \] Given that \( |A^3| = 125 \), we can substitute and solve for \( |A| \): \[ |A|^3 = 125 \] Taking the cube root of both sides: \[ |A| = \sqrt[3]{125} = 5 \]

Step 3: Solve for \( k \)
From the previous step, we have \( |A| = 5 \). So: \[ k^2 - 4 = 5 \] Solving for \( k \): \[ k^2 = 9 \quad \Rightarrow \quad k = 3 \text{ or } k = -3 \]

Conclusion:
The value of \( k \) is either \( 3 \) or \( -3 \).