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 :

We have, $P^2=I-P$ <br/><br/>$$ \begin{aligned} \Rightarrow P^4 & =(I-P)^2=I+P^2-2 P \\\\ & =2 I-3 P \text { [Using Eq. (i)] } \end{aligned} $$ <br/><br/>$$ \begin{aligned} \Rightarrow P^8 & =(2 I-3 P)^2 \\\\ & =4 I+9 P^2-12 P \\\\ & =13 I-21 P \text { [Using Eq. (i)] } \end{aligned} $$ <br/><br/>and <br/><br/>$$ \begin{aligned} P^6 & =(I-P)(2 I-3 P) \\\\ & =2 I-3 P-2 P+3 P^2 \\\\ & =5 I-8 P \text { [Using Eq. (i)] } \end{aligned} $$ <br/><br/>$\begin{aligned} & \text { Now, } P^8+P^6=18 I-29 P \\\\ & \text { and } P^8-P^6=8 I-13 P \\\\ & \therefore \alpha=8, \beta=6, \gamma=18, \delta=8 \\\\ & \therefore \alpha+\beta+\gamma-\delta=8+6+18-8=24\end{aligned}$