If $A =\begin{bmatrix}3&1\\-1&2\end{bmatrix}$ and $I =\begin{bmatrix}1&0\\0&1\end{bmatrix}$, then the value of $A^2 - 5A+ 6I$ is

$\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$

The correct answer is Option (1) → $\begin{bmatrix}-1&0\\0&-1\end{bmatrix}$

Given:

$A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$,

$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Find: $A^2 - 5A + 6I$

Calculate $A^2$:

$A^2 = A \cdot A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 \times 3 + 1 \times (-1) & 3 \times 1 + 1 \times 2 \\ -1 \times 3 + 2 \times (-1) & -1 \times 1 + 2 \times 2 \end{bmatrix} = \begin{bmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

Calculate $-5A$:

$-5A = -5 \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} -15 & -5 \\ 5 & -10 \end{bmatrix}$

Calculate $6I$:

$6I = 6 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$

Sum all terms:

$A^2 - 5A + 6I = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} + \begin{bmatrix} -15 & -5 \\ 5 & -10 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix} = \begin{bmatrix} 8 - 15 + 6 & 5 - 5 + 0 \\ -5 + 5 + 0 & 3 - 10 + 6 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$