If I- 1 0- 0 1 - and E - - 0 1- 0 0 - then 2I-3E 3-

The correct option is C 8I +36E Consider, 2I+3E 2[ 1 amp;0; 0 amp;1 ]+3[ 0 amp;1; 0 amp;0 ] ⇒[ 2 amp;0; 0 amp;2 ]+[ ...

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Byju's Answer

Standard XII

Mathematics

Equality of Matrices

If I=[ 1 0;...

Question

If $I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ and E = $[\begin{matrix} 0 & 1 0 & 0 \end{matrix}]$ , then ${(2 I + 3 E)}^{3} =$

8 I + 18 E

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4 I + 36 E

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8 I + 36 E

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2 I + 3 E

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Solution

The correct option is C $8 I + 36 E$
Consider, $2 I + 3 E$
$2 [\begin{matrix} 1 & 0 0 & 1 \end{matrix}] + 3 [\begin{matrix} 0 & 1 0 & 0 \end{matrix}]$
$\Rightarrow [\begin{matrix} 2 & 0 0 & 2 \end{matrix}] + [\begin{matrix} 0 & 3 0 & 0 \end{matrix}]$
$\Rightarrow [\begin{matrix} 2 & 3 0 & 2 \end{matrix}]$
Now $(2 I + 3 E)^{3}$
$\Rightarrow [\begin{matrix} 2 & 3 0 & 2 \end{matrix}] [\begin{matrix} 2 & 3 0 & 2 \end{matrix}] [\begin{matrix} 2 & 3 0 & 2 \end{matrix}]$
$\Rightarrow [\begin{matrix} 4 & 6 + 6 0 & 4 \end{matrix}] [\begin{matrix} 2 & 3 0 & 2 \end{matrix}]$
$= [\begin{matrix} 8 & 12 + 24 0 & 8 \end{matrix}] = [\begin{matrix} 8 & 36 0 & 8 \end{matrix}] = 8 I + 36 E$
$∴ (C)$

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If I=[1001] and E =[0100], then (2I+3E)3=

The correct option is C 8I+36EConsider, 2I+3E2[1001]+3[0100]⇒[2002]+[0300]⇒[2302]Now (2I+3E)3⇒[2302][2302][2302]⇒[46+604][2302]=[812+2408]=[83608]=8I+36E∴(C)

If $I = [\begin{matrix} 1 & 0 0 & 1 \end{matrix}]$ and E = $[\begin{matrix} 0 & 1 0 & 0 \end{matrix}]$ , then ${(2 I + 3 E)}^{3} =$