Question

If $A=[\begin{array}{cc}1& -1\\ 2& 3\end{array}]$ then show that $A^2-4A+5I_2=O$.

Answer 1

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

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

$=\left[\begin{array}{cc}1-2& -1-3\\ 2+6& -2+9\end{array}\right]$

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

LHS$=A^2-4A+5I=\left[\begin{array}{cc}-1& -4\\ 8& 7\end{array}\right]-\left[\begin{array}{cc}4& -4\\ 8& 12\end{array}\right]+\left[\begin{array}{cc}5& 0\\ 0& 5\end{array}\right]$

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

Hence proved