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Question

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


A
aI+bE
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B
a3I+b3E
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C
a3I+3ab2E
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D
a3I+3a2bE
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Solution

The correct option is B $$a^{3}I+3a^{2}bE$$
$$(aI +  bE)(aI + bE)(aI + bE)$$
$$(a^{2}I^{2} + b^{2}E^{2} + ab(IE + EI))(aI + bE)$$
$$I^{2} = I$$
$$E^{2} =\begin{bmatrix}
0 &1 \\
0 &0
\end{bmatrix}
\begin{bmatrix}
0 &1 \\
0 &0
\end{bmatrix} = \begin{bmatrix}
0 &0 \\
0 &0
\end{bmatrix}$$
$$E^{2}E = \begin{bmatrix}
0 &0 \\
0 &0
\end{bmatrix}$$
$$IE = EI = \begin{bmatrix}
0 &1 \\
0 &0
\end{bmatrix}= E$$
$$= a^{3}I + 3a^{2}bE$$

Mathematics

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