Question

If $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]$, show that $A^2-5A+7I=0$. Hence, find $A^{-1}$

Answer 1

Given $A=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right],A^2=AA=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]=\left[\begin{array}{cc}9-1& 3+2\\ -3-2& -1+4\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]$
Now, $A^2-5A+7I=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-5\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right],+7\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}8& 5\\ -5& 3\end{array}\right]-\left[\begin{array}{cc}15& 5\\ -5& 10\end{array}\right]+\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$
$=\left[\begin{array}{cc}8-15+7& 5-5+0\\ -5+5+0& 3-10+7\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]=0$
$\therefore A^2-5A+7I=0$
$\because |A|=\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right]=6+1=7\ne 0\therefore A^{-1}$ exists.
Now, $A.A-5A=-7I$
Multiply by $A^{-1}$ on both sides, we get
$A.A\left(A^{-1}\right)-5A{A}^{-1}=-7I{A}^{-1}$
$\Rightarrow AI-5I=7A^{-1}$ (using $A{A}^{-1}=IandI{A}^{-1}=A^{-1}$)
$\Rightarrow A^{-1}=-\frac{1}{7}(A-5I)\Rightarrow A^{-1}=\frac{1}{7}(5I-A)$
$=\frac{1}{7}(\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]-\left[\begin{array}{cc}3& 1\\ -1& 2\end{array}\right])=\frac{1}{7}\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\therefore A^{-1}=\frac{1}{7}\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]$
Note: When we multiply by $A^{-1}$ in the given equation, it is necessary that $|A|\ne 0$, If $|A|=0$, then $A^{-1}$ does not exist