Question

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

Answer 1

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

To Prove : $A^2-5A+7I=O$

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

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

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

Consider , $A^2-5A+7I$

$=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

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

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

$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

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

$A^{-1}$ :

Multiplying eqn (1) by $A^{-1}$

$A^2{A}^{-1}-5A{A}^{-1}+7I{A}^{-1}=O.A^{-1}$

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

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

$=5\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]-\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$

$=\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]-\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$

$=\left[\begin{array}{cc}5-3& 0-1\\ 0-(-1)& 5-2\end{array}\right]$

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

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