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If $\mathrm{A}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$, then what is $\mathrm{A}^{\mathrm{n}}$ equal to?
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Verified Answer
The correct answer is:
$\left[\begin{array}{ll}2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1} \\ 2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1}\end{array}\right]$
Given matrix is:
$\mathrm{A}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$
$\mathrm{A}^{2}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{ll}4+4 & 4+4 \\ 4+4 & 4+4\end{array}\right]=\left[\begin{array}{ll}2^{3} & 2^{3} \\ 2^{3} & 2^{3}\end{array}\right]$
$\begin{aligned} \mathrm{A}^{3} &=\left[\begin{array}{ll}8 & 8 \\ 8 & 8\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{ll}16+16 & 16+16 \\ 16+16 & 16+16\end{array}\right] \\ &=\left[\begin{array}{ll}32 & 32 \\ 32 & 32\end{array}\right]=\left[\begin{array}{cc}2^{5} & 2^{5} \\ 2^{5} & 2^{5}\end{array}\right] \end{aligned}$
Going this way we get
$\mathrm{A}^{4}=\left[\begin{array}{cc}2^{7} & 2^{7} \\ 2^{7} & 2^{7}\end{array}\right] \Rightarrow \mathrm{A}^{\mathrm{n}}=\left[\begin{array}{cc}2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1} \\ 2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1}\end{array}\right]$
$\mathrm{A}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]$
$\mathrm{A}^{2}=\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{ll}4+4 & 4+4 \\ 4+4 & 4+4\end{array}\right]=\left[\begin{array}{ll}2^{3} & 2^{3} \\ 2^{3} & 2^{3}\end{array}\right]$
$\begin{aligned} \mathrm{A}^{3} &=\left[\begin{array}{ll}8 & 8 \\ 8 & 8\end{array}\right]\left[\begin{array}{ll}2 & 2 \\ 2 & 2\end{array}\right]=\left[\begin{array}{ll}16+16 & 16+16 \\ 16+16 & 16+16\end{array}\right] \\ &=\left[\begin{array}{ll}32 & 32 \\ 32 & 32\end{array}\right]=\left[\begin{array}{cc}2^{5} & 2^{5} \\ 2^{5} & 2^{5}\end{array}\right] \end{aligned}$
Going this way we get
$\mathrm{A}^{4}=\left[\begin{array}{cc}2^{7} & 2^{7} \\ 2^{7} & 2^{7}\end{array}\right] \Rightarrow \mathrm{A}^{\mathrm{n}}=\left[\begin{array}{cc}2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1} \\ 2^{2 \mathrm{n}-1} & 2^{2 \mathrm{n}-1}\end{array}\right]$
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