Question

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

Then $(aI+bE{)}^3=a^{3}I+k{a}^{2}bE$

Find the value of $k.$

• A. 1
• B. 2
• C. 5
• D. 3

Answer 1

The correct option is D 3

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

and $(aI+bE{)}^3=a^{3}I+k{a}^{2}bE$

$E^2=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

$\Rightarrow E^2=0$ and $E^3=0$

Now,

$(aI+bE{)}^3=a^3{I}^3+b^3{E}^3+3abE(aI+bE)$

$=a^{3}I+b^3{E}^3+3a^{2}bE+3a{b}^2{E}^2$

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

$\therefore k=3$