Let a square matrix such that <img alt="\mathrm{P}^2=\mathrm{I}-\mathrm{P}" src="https:

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

Option: 1
40

Option: 2
22

Option: 3
24

Option: 4
18

Answers (1)

$\mathrm{P}^2=\mathrm{I}-\mathrm{P} \\$
$\mathrm{P}^\alpha+\mathrm{P}^\beta=\gamma \mathrm{I}-29 \mathrm{P} \\$
$\mathrm{P}^\alpha-\mathrm{P}^\beta=\delta \mathrm{I}-13 \mathrm{P} \\$
$\mathrm{P}^4=(\mathrm{I}-\mathrm{P})^2=\mathrm{I}+\mathrm{P}^2-2 \mathrm{P} \\$
$\mathrm{P}^4=\mathrm{I}+\mathrm{I}-\mathrm{P}-2 \mathrm{P}=2 \mathrm{I}-3 \mathrm{P} \\$
$\mathrm{P}^8=\left(\mathrm{P}^4\right)^2=(2 \mathrm{I}-3 \mathrm{P})^2 \quad=4 \mathrm{I}+9 \mathrm{P}^2-12 \mathrm{P} \\$
                                                     $\quad=4 \mathrm{I}+9(\mathrm{I}-\mathrm{P})-12 \mathrm{P} \\$
                                                     $\mathrm{P}^8=13 \mathrm{I}-21 \mathrm{P} \\$ ..........(1)
$\mathrm{P}^6=\mathrm{P}^4 \cdot \mathrm{P}^2 \quad=(2 \mathrm{I}-3 \mathrm{P})(\mathrm{I}-\mathrm{P})$
                              $=2 \mathrm{I}-5 \mathrm{P}+3 \mathrm{P}^2$
                              $=2 \mathrm{I}-5 \mathrm{P}+3(\mathrm{I}-\mathrm{P})$
                              $=5 \mathrm{I}-8 \mathrm{P}$ ................................(2)
$(1)+(2) \: \: \: \: \: \: \: \: \: \: \: (1)-(2)$
$\mathrm{P}^8-\mathrm{P}^6=18 \mathrm{I}-29 \mathrm{P}$                             $\mathrm{P}^8-\mathrm{P}^6=8 \mathrm{I}-13 \mathrm{P}$

From (A)