Let P be a square matrix such that P^2=I-P.

Question

Let $P$ be a square matrix such that $P^{2}=I-P$. For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$P^{\alpha}+P^{\beta}=\gamma I-29 P$$ and $$P^{\alpha}-P^{\beta}=\delta I-13 P$$, then $$\alpha+\beta+\gamma-\delta$$ is equal to :

Accepted Answer

We have, $P^2=I-P$

$$
\begin{aligned}
\Rightarrow P^4 & =(I-P)^2=I+P^2-2 P \\
& =2 I-3 P \text { [Using Eq. (i)] }
\end{aligned}
$$

$$
\begin{aligned}
\Rightarrow P^8 & =(2 I-3 P)^2 \\
& =4 I+9 P^2-12 P \\
& =13 I-21 P \text { [Using Eq. (i)] }
\end{aligned}
$$

and

$$
\begin{aligned}
P^6 & =(I-P)(2 I-3 P) \\
& =2 I-3 P-2 P+3 P^2 \\
& =5 I-8 P \text { [Using Eq. (i)] }
\end{aligned}
$$

$\begin{aligned} & \text { Now, } P^8+P^6=18 I-29 P \\ & \text { and } P^8-P^6=8 I-13 P \\ & \therefore \alpha=8, \beta=6, \gamma=18, \delta=8 \\ & \therefore \alpha+\beta+\gamma-\delta=8+6+18-8=24\end{aligned}$

Solutions · 1