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Derivative of a Determinant

If I=[ 1 0;...

Question

If $I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] . E = [\begin{matrix} 0 & 1 0 & 0 \end{matrix}]$ then show that $(a I + b E)^{3} = a^{3} I + 3 a^{2} b E$ .

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Solution

$\begin{matrix} I = (\begin{matrix} 1 & 0 0 & 1 \end{matrix}) E = (\begin{matrix} 0 & 1 0 & 0 \end{matrix}) a I + b E = (\begin{matrix} a & 0 0 & a \end{matrix}) + (\begin{matrix} 0 & b 0 & 0 \end{matrix}) = (\begin{matrix} a & b 0 & a \end{matrix}) \Rightarrow {(a I + b E)}^{3} = (\begin{matrix} a & b 0 & a \end{matrix}) (\begin{matrix} a & b 0 & a \end{matrix}) (\begin{matrix} a & b 0 & a \end{matrix}) = (\begin{matrix} a^{3} & 3 a^{2} b 0 & a^{3} \end{matrix}) = (\begin{matrix} a^{3} & 0 0 & a^{3} \end{matrix}) + (\begin{matrix} 0 & 3 a^{2} b 0 & 0 \end{matrix}) = a^{3} (\begin{matrix} 1 & 0 0 & 1 \end{matrix}) + 3 a^{2} b (\begin{matrix} 1 & 0 0 & 1 \end{matrix}) {(a I + b E)}^{3} = a^{3} I + 3 a^{2} b E \end{matrix}$

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Similar questions

I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

E = [\begin{matrix} 0 & 1 0 & 0 \end{matrix}]

(a I + b E)^{3} = a^{3} I + k a^{2} b E

Find the value of

k .

I \equiv [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

E \equiv [\begin{matrix} 0 & 1 0 & 0 \end{matrix}]

{(a l + b E)}^{3} = a^{3} l + 3 a^{2} b E

, where I is unit matrix of order

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E = [\begin{matrix} 0 & 1 0 & 0 \end{matrix}],

(a I + b E)^{3} =

I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]

[\begin{matrix} 0 & 1 0 & 0 \end{matrix}]

{(2 I + 3 E)}^{3} =

I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}], E = [\begin{matrix} 0 & 1 0 & 0 \end{matrix}]

(a I + b E)^{3} =

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