If $I = [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}], E = [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}]$ , then ${(a I + b E)}^{3} =$

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$a I + b E$

$a^{3} I + b^{3} E$

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

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

answer is D.

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Detailed Solution

$a I = a [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$ = $[\begin{matrix} a & 0 \\ 0 & a \end{matrix}]$ and $b E = b [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] = [\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}]$ then

$(a I + b E) =$ $[\begin{matrix} a & 0 \\ 0 & a \end{matrix}]$ + $[\begin{matrix} 0 & b \\ 0 & 0 \end{matrix}]$ = $[\begin{matrix} a & b \\ 0 & a \end{matrix}]$

${(a I + b E)}^{3} =$ $[\begin{matrix} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{matrix}]$

$a^{3} I + 3 a^{2} b E = a^{3} [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}] + 3 a^{2} b [\begin{matrix} 0 & 1 \\ 0 & 0 \end{matrix}] = [\begin{matrix} a^{3} & 3 a^{2} b \\ 0 & a^{3} \end{matrix}]$