Question

If $A=\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]$ and $I$ is the units matrix of order $2$, then $A^2-4A+5$ equals to?

Answer 1

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

$\Rightarrow A^2-4A+5I$

Let $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]-4\left[\begin{array}{cc}1& -1\\ 2& 3\end{array}\right]+5\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}1\times 1-1\times 2& 1\times -1(-1)\times 3\\ 2\times 1+3\times 2& 2\times (-1)+3\times 3\end{array}\right]-\left[\begin{array}{cc}4& -4\\ 8& 12\end{array}\right]+\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]$

$=\left[\begin{array}{cc}-1& -4\\ 8& 7\end{array}\right]-\left[\begin{array}{cc}4& -4\\ 8& 12\end{array}\right]+\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]$

$=\left[\begin{array}{cc}-1-4+5& -4+4+0\\ 8-8+0& 7-12+5\end{array}\right]$

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

$=0$

Hence, the answer is $0.$