If A = […] , show that A^2 - 5A + 7I = O. Hence, find A^-1.

The correct answer is Option (2) → $\frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$ ##

$A^2 = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$

$5A = \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix}, \quad 7I = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix}$

$\Rightarrow A^2 - 5A + 7I = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$

To find $A^{-1}$:

$\Rightarrow A^{-1}(A^2 - 5A + 7I) = A^{-1}O$

$\Rightarrow A - 5I + 7A^{-1} = O$

$\Rightarrow 7A^{-1} = 5I - A$

$\Rightarrow A^{-1} = \frac{1}{7} \left( \begin{bmatrix} 5 & 0 \\ 0 & 5 \end{bmatrix} - \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \right)$

$\Rightarrow A^{-1} = \frac{1}{7} \begin{bmatrix} 2 & -1 \\ 1 & 3 \end{bmatrix}$