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

Given,

$P$ be a square matrix such that $P^2=I-P$,

Now solving $P^2=I-P$ we get,

$\Rightarrow P^4=\left(I-P\right)\left(I-P\right)$

$\Rightarrow P^4=I+P^2-2P$

$\Rightarrow P^4=I+I-P-2P=2I-3P$

Now solving, $P^4·P^2=P^6=2I-5P+3P^2$ 

$\Rightarrow P^6=2I-5P+3\left(I-P\right)=5I-8P$... $\left(\mathrm{i}\right)$

Similarly $P^8=13I-21P$ ... $\left(\mathrm{ii}\right)$

Now adding $\left(\mathrm{ii}\right)+\left(\mathrm{i}\right)$ we get,

$P^8+P^6=18I-29P$

Now subtracting $\left(\mathrm{ii}\right)-\left(\mathrm{i}\right)$ we get,

$P^8-P^6=8I-13P$

Now comparing with $P^{\alpha }-P^{\beta }=\delta l-13P \& P^{\alpha }+P^{\beta }=\gamma l-29P$ we get,

$\alpha =8, \beta =6, \gamma =18, \delta =8$

Hence, $\alpha +\beta +\gamma -\delta =8+6+18-8=24$