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

Given:

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

Now,we have to show that $A^2-5A+7I=0$(null matrix)

LHS,

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

$=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]-5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]=\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]$

$=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$ (null matrix ) (proved)

$A^{-1}=\frac{adj\left(A\right)}{\left|A\right|}A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]adj\left(A\right)=\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\left|A\right|=6+1=7\therefore A^{-1}=\frac{1}{7}\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]=\left[\begin{array}{cc}\frac{2}{7}& \frac{-1}{7}\\ \frac{1}{7}& \frac{3}{7}\end{array}\right]$

Note: if we have a $2\times 2$ matrix $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ its adjoint is $\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$