Question

Let \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), then \( A^2 - 5A + 6I = ? \)

• A. \( \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \)
• B. \( \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} \)
• C. \( \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \)
• D. \( \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \)

Hint:

For matrix polynomials, compute each term separately and use the Cayley-Hamilton theorem for verification if needed.

Answer 1

The Correct Option is B

- Given \( A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), compute \( A^2 - 5A + 6I \).
- First, compute \( A^2 \): \[ A^2 = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + 2 \cdot 3 & 1 \cdot 2 + 2 \cdot 4 \\ 3 \cdot 1 + 4 \cdot 3 & 3 \cdot 2 + 4 \cdot 4 \end{bmatrix} = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} \]
- Compute \( 5A \): \[ 5A = 5 \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} \]
- Compute \( 6I \), where \( I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \): \[ 6I = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} \]
- Now, compute \( A^2 - 5A + 6I \): \[ A^2 - 5A = \begin{bmatrix} 7 & 10 \\ 15 & 22 \end{bmatrix} - \begin{bmatrix} 5 & 10 \\ 15 & 20 \end{bmatrix} = \begin{bmatrix} 7 - 5 & 10 - 10 \\ 15 - 15 & 22 - 20 \end{bmatrix} = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \] \[ A^2 - 5A + 6I = \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 2 + 6 & 0 \\ 0 & 2 + 6 \end{bmatrix} = \begin{bmatrix} 8 & 0 \\ 0 & 8 \end{bmatrix} \]