Question

If $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right].E=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ then show that $(aI+bE{)}^3=a^{3}I+3a^{2}bE$.

Answer 1

$\begin{array}{c}I=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)E=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)\\ aI+bE=\left(\begin{array}{cc}a& 0\\ 0& a\end{array}\right)+\left(\begin{array}{cc}0& b\\ 0& 0\end{array}\right)\\ =\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)\\ \Rightarrow {\left(aI+bE\right)}^3=\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)\left(\begin{array}{cc}a& b\\ 0& a\end{array}\right)\\ =\left(\begin{array}{cc}{a}^3& 3a^{2}b\\ 0& a^3\end{array}\right)\\ =\left(\begin{array}{cc}{a}^3& 0\\ 0& a^3\end{array}\right)+\left(\begin{array}{cc}0& 3a^{2}b\\ 0& 0\end{array}\right)\\ =a^3\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)+3a^{2}b\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\\ {\left(aI+bE\right)}^3=a^{3}I+3a^{2}bE\end{array}$