# Solve this

Question:

If $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$, show that $A^{2}-5 A+7 I=O$. Hence, find $A^{-1}$.

Solution:

$A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$

$\Rightarrow A^{2}=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$

and

$A^{2}-5 A+7 I=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$

$\Rightarrow A^{2}-5 A+7 I=\left[\begin{array}{cc}8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=O$

Now,

$A^{2}-5 A+7 I=0$

$\Rightarrow A^{2}-5 A=-7 I$

$\Rightarrow A^{-1} A^{2}-5 A^{-1} A=-7 A^{-1} I \quad\left[\right.$ Pre $-$ multiplying both sides by $\left.A^{-1}\right]$

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

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

$\Rightarrow A^{-1}=-\frac{1}{7}\left\{\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]-\left[\begin{array}{ll}5 & 0 \\ 0 & 5\end{array}\right]\right\}=\frac{1}{7}\left[\begin{array}{cc}2 & -1 \\ 1 & 3\end{array}\right]$