Question

If $I\equiv \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ and $E\equiv \left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$ then show that ${(al+bE)}^3=a^{3}l+3a^{2}bE$, where I is unit matrix of order $2$.

Answer 1

$I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],E=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$

${(aI+bE)}^3\Rightarrow aI=\left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right],bE\left[\begin{array}{cc}0& b\\ 0& 0\end{array}\right]$

$\Rightarrow aI+bE=\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right]$

${(aI+bE)}^2=\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right]=\left[\begin{array}{cc}{a}^2& ab+ab\\ 0& a^2\end{array}\right]=\left[\begin{array}{cc}{a}^2& 2ab\\ 0& a^2\end{array}\right]$

${(aI+bE)}^3=\left[\begin{array}{cc}{a}^2& 2ab\\ 0& a^2\end{array}\right]\left[\begin{array}{cc}a& b\\ 0& a\end{array}\right]=\left[\begin{array}{cc}{a}^3& a^2+2a^{2}b\\ 0& a^3\end{array}\right]$

$=\left[\begin{array}{cc}{a}^3& 3a^{2}b\\ 0& a^3\end{array}\right]=L.H.S$

$R.H.S=a^{3}I+3a^{2}bE$

$\Rightarrow a^{3}I=\left[\begin{array}{cc}{a}^3& 0\\ 0& a^3\end{array}\right],3a^{2}bE=\left[\begin{array}{cc}0& 3a^{2}b\\ 0& 0\end{array}\right]$

$a^{3}I+3a^{2}bE=\left[\begin{array}{cc}{a}^3& 3a^{2}b\\ 0& a^3\end{array}\right]=R.H.S$

$\therefore L.H.S=R.H.S$