Question

A square matrix $P$ satisfies $P^2 = I - P,$ where $I$ is the identity matrix. If $P^n = 5I - 8P$, then $n$ is equal to

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

Answer: (C)

$\because P^{3}=P\left(I-P\right)$
$= PI - P^{2} = PI - \left(I - P\right)$
$= P - I + P = 2P - 1 $
Now, $P^{4}=P.P^3 $
$\Rightarrow P^4 = P ( 2P -I )$
$\Rightarrow P^{4}=2P^{2}-P$
$\Rightarrow P^{4}=2I-2P-P$
$\Rightarrow P^{4}=2I-3P$
and $P^{5}=P\left(2I-3P\right)$
$\Rightarrow P^{5}=2P-3\left(I-P\right)$
$\Rightarrow P^{5}=5P-3I$
Also, $P^{6}=P\left(5P-3I\right)$
$\Rightarrow P^{6}=5P^{2}-3P$
$\Rightarrow P^{6}=5\left(I-P\right)-3P$
$\Rightarrow P^{6}=5I-8P$
So, $n=6$