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Question

 If  $$I=\begin{bmatrix}
1 & 0\\
0 & 1
\end{bmatrix}$$ and E =$$\begin{bmatrix}
0 & 1\\
0 & 0
\end{bmatrix}$$, then $$\left ( 2I+3E \right )^{3}=$$ 


A
8I+18E
loader
B
4I+36E
loader
C
8I+36E
loader
D
2I+3E
loader

Solution

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

Mathematics

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