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Question: Answered & Verified by Expert
If $P=\left(\begin{array}{rrr}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right),$ then $P^{5}$ is equal to
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Options:
  • A $P$
  • B $2 P$
  • C $-P$
  • D $-2 P$
Solution:
2508 Upvotes Verified Answer
The correct answer is: $P$
Given, $P=\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right)$
$\begin{aligned} P^{2} &=P \cdot P=\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right)\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right) \\ &=\left(\begin{array}{ccc}4+2-4 & -4-6+8 & -8 & -8+12 \\ -2-3+4 & 2+9-8 & 4+12-12 \\ 2+2-3 & -2-6+6 & -4-8+9\end{array}\right) \\ &=\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right)=P \\ \therefore P^{4}=P^{2}=P \\ \Rightarrow P^{5}=P^{2}=P \end{aligned}$

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