Question

If \[ A = \begin{bmatrix} 1 & 0 & 2\\ 2 & 1 & 3 \\3 & 2 & 4 \end{bmatrix}, \] then evaluate \( A^2 - 5A + 6I \)=

• A. \( \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 12 \end{bmatrix} \)
• B. \( \begin{bmatrix} 8 & 4 & 0 \\ 3 & 6 & 4 \\ 4 & 0 & 14 \end{bmatrix} \)
• C. \( \begin{bmatrix} 8 & 6 & 0 \\ 3 & 8 & 4 \\ 2 & 0 & 14 \end{bmatrix} \)
• D. \( \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix} \)

Hint:

To compute matrix expressions like \( A^2 - 5A + 6I \), first determine \( A^2 \), then scale \( A \) and \( I \), and finally perform matrix addition and subtraction.

Answer 1

The Correct Option is D

We are given the matrix: \[ A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix}. \] 
Step 1: Compute \( A^2 \) \[ A^2 = A \times A = \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} \cdot \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix}. \] 
Performing matrix multiplication: \[ A^2 = \begin{bmatrix} (1 \times 1 + 0 \times 2 + 2 \times 3) & (1 \times 0 + 0 \times 1 + 2 \times 2) & (1 \times 2 + 0 \times 3 + 2 \times 4) \\ (2 \times 1 + 1 \times 2 + 3 \times 3) & (2 \times 0 + 1 \times 1 + 3 \times 2) & (2 \times 2 + 1 \times 3 + 3 \times 4) \\ (3 \times 1 + 2 \times 2 + 4 \times 3) & (3 \times 0 + 2 \times 1 + 4 \times 2) & (3 \times 2 + 2 \times 3 + 4 \times 4) \end{bmatrix} \] \[ = \begin{bmatrix} 1 + 0 + 6 & 0 + 0 + 4 & 2 + 0 + 8 \\ 2 + 2 + 9 & 0 + 1 + 6 & 4 + 3 + 12 \\ 3 + 4 + 12 & 0 + 2 + 8 & 6 + 6 + 16 \end{bmatrix} \] \[ = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\ 19 & 10 & 28 \end{bmatrix}. \] 
Step 2: Compute \( 5A \) \[ 5A = 5 \times \begin{bmatrix} 1 & 0 & 2 \\ 2 & 1 & 3 \\ 3 & 2 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix}. \] 
Step 3: Compute \( 6I \) \[ 6I = 6 \times \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\0 & 0 & 6 \end{bmatrix}. \] 
Step 4: Compute \( A^2 - 5A + 6I \) \[ A^2 - 5A + 6I = \begin{bmatrix} 7 & 4 & 10 \\ 13 & 7 & 19 \\19 & 10 & 28 \end{bmatrix} - \begin{bmatrix} 5 & 0 & 10 \\ 10 & 5 & 15 \\ 15 & 10 & 20 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}. \] \[ = \begin{bmatrix} (7 - 5 + 6) & (4 - 0 + 0) & (10 - 10 + 0) \\ (13 - 10 + 0) & (7 - 5 + 6) & (19 - 15 + 0) \\ (19 - 15 + 0) & (10 - 10 + 0) & (28 - 20 + 6) \end{bmatrix} \] \[ = \begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix}. \] 
Thus, the correct answer is: \[ \boxed{\begin{bmatrix} 8 & 4 & 0 \\ 3 & 8 & 4 \\ 4 & 0 & 14 \end{bmatrix}} \]