Question

If $a,b,c,d$ are real numbers and $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$
Then $A^2-(a+d)A+(ad-bc)I=\left[\begin{array}{cc}k& m\\ n& p\end{array}\right]$. Find the value of $k+m+n+p$ ?

Answer 1

Given that $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$\therefore A^2=A.A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

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

$\therefore A^2-(a+d)A+(ad-bc)I=\left[\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]-(a+d)\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]+(ad-bc)\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$=\left[\begin{array}{cc}{a}^2+bc& ab+bd\\ ac+cd& bc+d^2\end{array}\right]-\left[\begin{array}{cc}(a+d)a& (a+d)b\\ (a+d)c& (a+d)d\end{array}\right]+\left[\begin{array}{cc}ad-bc& 0\\ 0& ad-bc\end{array}\right]$

$=\left[\begin{array}{cc}{a}^2+bc-a^2-ad+ad-bc& ab+bd-ab-bd+0\\ ac+cd-ac-cd+0& bc+d^2-ad-d^2+ad-bc\end{array}\right]$

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

$=O$

$\therefore m+k+n+p=0$