If P is a square matrix with P^2=P and if I i | vedclass

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(D) Given that $P^2 = P$. This implies that $P$ is an idempotent matrix.We need to evaluate $(P+I)^4$.Using the binomial expansion theorem for matrice

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$I+15P$

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(D) Given that $P^2 = P$. This implies that $P$ is an idempotent matrix.We need to evaluate $(P+I)^4$.Using the binomial expansion theorem for matrices,since $P$ and $I$ commute $(PI = IP = P)$:$(P+I)^n = \sum_{k=0}^{n} \binom{n}{k} P^k I^{n-k}$.For $n=4$:$(P+I)^4 = \binom{4}{0} P^0 I^4 + \binom{4}{1} P^1 I^3 + \binom{4}{2} P^2 I^2 + \binom{4}{3} P^3 I^1 + \binom{4}{4} P^4 I^0$.Since $I^n = I$ and $P^k = P$ for all $k \ge 1$ (because $P^2=P, P^3=P^2 \cdot P = P \cdot P = P$,etc.):$(P+I)^4 = I + 4P + 6P + 4P + P$.$(P+I)^4 = I + (4+6+4+1)P$.$(P+I)^4 = I + 15P$.

If $P$ is a square matrix with $P^2=P$ and if $I$ is the unit matrix of the same order as of $P$,then $(P+I)^4=$

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