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If the matrix $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right],$ then
$A^{n}=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{array}\right], n \in N,$ where
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$A^{n}=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{array}\right], n \in N,$ where
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Verified Answer
The correct answer is:
$a=2^{n}, b=n 2^{n}$
We have. $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right]$
$\therefore \quad A^{2}=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right]=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 8 & 0 & 4\end{array}\right]$
$=\left[\begin{array}{ccc}2^{2} & 0 & 0 \\ 0 & 2^{2} & 0 \\ 2 \cdot 2^{2} & 0 & 2^{2}\end{array}\right]$
Now, $\quad A^{n}=\left[\begin{array}{ccc}2^{n} & 0 & 0 \\ 0 & 2^{n} & 0 \\ n-2^{n} & 0 & 2^{n}\end{array}\right]$
$A^{n}=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{array}\right]$
$\therefore \quad\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{array}\right]=\left[\begin{array}{ccc}2^{n} & 0 & 0 \\ 0 & 2^{n} & 0 \\ n \cdot 2^{n} & 0 & 2^{n}\end{array}\right]$
Hence, $\quad a=2^{n}$
and $\quad b=n 2^{n}$
$\therefore \quad A^{2}=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right]=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 8 & 0 & 4\end{array}\right]$
$=\left[\begin{array}{ccc}2^{2} & 0 & 0 \\ 0 & 2^{2} & 0 \\ 2 \cdot 2^{2} & 0 & 2^{2}\end{array}\right]$
Now, $\quad A^{n}=\left[\begin{array}{ccc}2^{n} & 0 & 0 \\ 0 & 2^{n} & 0 \\ n-2^{n} & 0 & 2^{n}\end{array}\right]$
$A^{n}=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{array}\right]$
$\therefore \quad\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{array}\right]=\left[\begin{array}{ccc}2^{n} & 0 & 0 \\ 0 & 2^{n} & 0 \\ n \cdot 2^{n} & 0 & 2^{n}\end{array}\right]$
Hence, $\quad a=2^{n}$
and $\quad b=n 2^{n}$
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