Let P be an m times m matrix such that P^2=P. | vedclass

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(D) Given that $P^2 = P$. This is an idempotent matrix.We use the binomial expansion for $(I+P)^n$:$(I+P)^n = \sum_{k=0}^{n} \binom{n}{k} I^{n-k} P^k$

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$I+(2^n-1)P$

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(D) Given that $P^2 = P$. This is an idempotent matrix.We use the binomial expansion for $(I+P)^n$:$(I+P)^n = \sum_{k=0}^{n} \binom{n}{k} I^{n-k} P^k$Since $I^m = I$ for any $m \ge 1$ and $P^k = P$ for all $k \ge 1$ (because $P^2=P, P^3=P^2P=PP=P$,etc.),we have:$(I+P)^n = I + \sum_{k=1}^{n} \binom{n}{k} P$$(I+P)^n = I + P \left( \sum_{k=1}^{n} \binom{n}{k} \right)$We know that $\sum_{k=0}^{n} \binom{n}{k} = 2^n$,so $\sum_{k=1}^{n} \binom{n}{k} = 2^n - \binom{n}{0} = 2^n - 1$.Therefore,$(I+P)^n = I + (2^n - 1)P$.

Let $P$ be an $m \times m$ matrix such that $P^2=P$. Then,$(I+P)^n$ equals

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