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

Step 2: Determinant of \( A^3 \)
The property \( \det(A^n) = (\det A)^n \) for any square matrix \( A \) is applied. Thus, \( \det(A^3) = (\det A)^3 = 6^3 \). Calculation yields \( 6^3 = 6 \times 6 \times 6 = 216 \).

Step 3: Alternative approach
Compute \( A^3 \). First, \( A^2 = A \cdot A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} \cdot \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 0 & 9 \end{bmatrix} \). Then, \( A^3 = A^2 \cdot A = \begin{bmatrix} 4 & 0 \\ 0 & 9 \end{bmatrix} \cdot \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 0 & 27 \end{bmatrix} \). The determinant of \( A^3 \) is \( \det(A^3) = 8 \times 27 = 216 \). Both methods yield consistent results.

Step 4: Option verification
Option (1) which is 216 aligns with the computed result.