Question

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

Answer 1

Given that:

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],$ $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

To prove:

$A^2-(a+d)A=(bc-ad)I$

Solution:

$A^2=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\times \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]$

$(a+d)A=(a+d)\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=\left[\begin{array}{cc}{a}^2+ad& ab+bd\\ ac+cd& ad+d^2\end{array}\right]$

$A^2-(a+d)A=\left[\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]-\left[\begin{array}{cc}{a}^2+ad& ab+bd\\ ac+cd& ad+d^2\end{array}\right]$

$=\left[\begin{array}{cc}{a}^2+bc-a^2-ad& ab+bd-ab-bd\\ ac+cd-ac-cd& bc+d^2-ad-d^2\end{array}\right]$

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

$(bc-ad)I=(bc-ad)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

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

Hence, $A^2-(a+d)A=(bc-ad)I$