Question

Let \( A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \). The determinant of \( A^3 \) is:

• A. 216
• B. 27
• C. 8
• D. 1

Hint:

The determinant of matrix power is the determinant raised to that power: \( \det(A^n) = (\det A)^n \)

Answer 1

The Correct Option is A

Step 1: Determinant of matrix \( A \)
For a 2x2 matrix \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), the determinant is \( ad - bc \). For: \[ A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \] \[ \det A = (2)(3) - (0)(0) = 6 \] Since \( A \) is diagonal, the determinant is the product of diagonal elements: \( 2 \times 3 = 6 \).

Step 2: Determinant of \( A^3 \)
Using the property of determinants, for any square matrix \( A \): \[ \det(A^n) = (\det A)^n \] So: \[ \det(A^3) = (\det A)^3 = 6^3 \] Calculate: \[ 6^3 = 6 \times 6 \times 6 = 216 \]
Step 3: Alternative approach
Compute \( A^3 \): \[ A^2 = A \cdot A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 9 \end{bmatrix} \] \[ A^3 = A^2 \cdot A = \begin{bmatrix} 4 & 0 \\ 0 & 9 \end{bmatrix} \cdot \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 0 & 27 \end{bmatrix} \] Determinant of \( A^3 \): \[ \det(A^3) = 8 \times 27 = 216 \] Both methods agree.

Step 4: Check options
Option (1) 216 matches our result.