Question

Let $P$ be a non-singular matrix such that $I+P+P^2+\cdots +P^n=O$. Then $P^{-1}$ is equal to

• A. $P$
• B. $-P^n$
• C. $P^{n-1}$
• D. $P^n$

Answer 1

The correct option is D $P^n$

$I+P+P^2+\cdots +P^n=O\cdots \left(1\right)$
Since, $P$ is non-singular, $P^{-1}$ exists.
Multiplying both sides by $P^{-1}$, we get
$P^{-1}+I+IP+\cdots +I{P}^{n-1}=O{P}^{-1}$
$\Rightarrow P^{-1}+I(I+P+\cdots +P^{n-1})=O$
$\Rightarrow P^{-1}=-I(I+P+P^2+\cdots +P^{n-1})$
$\Rightarrow P^{-1}=-(-P^n)\left[\text{From}\right(1\left)\right]$
$\Rightarrow P^{-1}=P^n$