Question

If $A=(\begin{array}{cc}a& b\\ c& d\end{array})$ and $I_2=(\begin{array}{cc}1& 0\\ 0& 1\end{array})$, then show that $A^2-(a+d)A=(bc-ad)I_2$.

Answer 1

Consider $A^2=A\times A$
$=(\begin{array}{cc}a& b\\ c& d\end{array})(\begin{array}{cc}a& b\\ c& d\end{array})=(\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array})$ ............. (1)
Now, $(a+d)A=(a+d)(\begin{array}{cc}a& b\\ c& d\end{array})$
$=(\begin{array}{cc}{a}^2+ad& ab+bd\\ ac+cd& ad+d^2\end{array})$ ............. (2)
From (1) and (2) we get,
$A^2-(a+d)A=(\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array})-(\begin{array}{cc}{a}^2+ad& ab+bd\\ ac+cd& ad+d^2\end{array})$
$=(\begin{array}{cc}bc-ad& 0\\ 0& bc-ad\end{array})=(bc-ad)(\begin{array}{cc}1& 0\\ 0& 1\end{array})$
Thus, $A^2-(a+d)A=(bc-ad)I_2$.