Let P be a square matrix such that P^2=I-P. For α ,β ,γ ,δ ∈ mathbbN, if P^α +P^β =γ l-29P and P^α -P^β =δ l-13P, then α +β +γ -δ is equal to

Given,

$P$P be a square matrix such that ${P}^{2}=I-P$P2=I−P,

Now solving ${P}^{2}=I-P$P2=I−P we get,

$$\Rightarrow {P}^{4}=(I-P)(I-P)$$⇒P4=(I−P)(I−P)

$$\Rightarrow {P}^{4}=I+{P}^{2}-2P$$⇒P4=I+P2−2P

$$\Rightarrow {P}^{4}=I+I-P-2P=2I-3P$$⇒P4=I+I−P−2P=2I−3P

Now solving, ${P}^{4}\cdot {P}^{2}={P}^{6}=2I-5P+3{P}^{2}$P4⋅P2=P6=2I−5P+3P2

$\Rightarrow {P}^{6}=2I-5P+3(I-P)=5I-8P$⇒P6=2I−5P+3(I−P)=5I−8P... $(i)$(i)

Similarly ${P}^{8}=13I-21P$P8=13I−21P ... $(\mathrm{ii})$(ii)

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

$${P}^{8}+{P}^{6}=18I-29P$$P8+P6=18I−29P

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

$${P}^{8}-{P}^{6}=8I-13P$$P8−P6=8I−13P

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

$$\alpha =8,\beta =6,\gamma =18,\delta =8$$α=8,β=6,γ=18,δ=8

Hence, $\alpha +\beta +\gamma -\delta =8+6+18-8=24$α+β+γ−δ=8+6+18−8=24