Question

lf $I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],E=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, then $(aI+bE{)}^3=$

• A. $aI+bE$
• B. $a^{3}I+b^{3}E$
• C. $a^{3}I+3a{b}^{2}E$
• D. $a^{3}I+3a^{2}bE$

Answer 1

Answer: (D)

$a^{3}I+3a^{2}bE$

$(aI+bE)(aI+bE)(aI+bE)$
$(a^2{I}^2+b^2{E}^2+ab(IE+EI\left)\right)(aI+bE)$
$I^2=I$
$E^2=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$E^{2}E=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
$IE=EI=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=E$
$=a^{3}I+3a^{2}bE$