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If $\mathrm{P}$ is a non-singular matrix such that $\mathrm{I}+\mathrm{P}+\mathrm{P}^2 \ldots+\mathrm{P}^{\mathrm{n}}=$ $0(0$ denotes the null matrix $)$, then $\mathrm{P}^{-1}=$
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Verified Answer
The correct answer is:
$\mathrm{P}^{\mathrm{n}}$
$I+P+P^2+\ldots . .+P^n=0... (i)$
$\left(I+P+P^2+\ldots . .+P^{n-1}\right)=0-P^n=-P^n$
Pre multiply (i) with $P^{-1}$
$\begin{aligned}
& P^{-1}+I+P+P^2+\ldots . .+P^{n-1}=0 \\
& P^{-1}=0-\left(I+P+P^2+\ldots .+P^{n-1}\right) \Rightarrow P^{-1}=P^n
\end{aligned}$
$\left(I+P+P^2+\ldots . .+P^{n-1}\right)=0-P^n=-P^n$
Pre multiply (i) with $P^{-1}$
$\begin{aligned}
& P^{-1}+I+P+P^2+\ldots . .+P^{n-1}=0 \\
& P^{-1}=0-\left(I+P+P^2+\ldots .+P^{n-1}\right) \Rightarrow P^{-1}=P^n
\end{aligned}$
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