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, $n=6$