Verified Solution • 2025–26 Session moderate

QJEEM23Ma1e7a: JEE Main 2023 Mathematics - Class 12 JEEM23Ma1e7a Solution: Let be a square matrix such that . For , if and , then is equal …

Let PP be a square matrix such that P2=I−PP^{2}=I-P . For α,β,γ,δ∈N\alpha, \beta, \gamma, \delta \in \mathbb{N} , if Pα+Pβ=γI−29PP^{\alpha}+P^{\beta}=\gamma I-29 P and Pα−Pβ=δI−13PP^{\alpha}-P^{\beta}=\delta I-13 P , then α+β+γ−δ\alpha+\beta+\gamma-\delta is equal to :

Multiple Choice Options

  • A 18
  • B 22
  • C 24
  • D 40

Step-by-Step Verified Solution

We have, P2=I−PP^2=I-P

⇒P4amp;=(I−P)2=I+P2−2Pamp;=2I−3P [Using Eq. (i)] \begin{aligned}\Rightarrow P^4 & =(I-P)^2=I+P^2-2 P \\\\& =2 I-3 P \text { [Using Eq. (i)] }\end{aligned}

⇒P8amp;=(2I−3P)2amp;=4I+9P2−12Pamp;=13I−21P [Using Eq. (i)] \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

P6amp;=(I−P)(2I−3P)amp;=2I−3P−2P+3P2amp;=5I−8P [Using Eq. (i)] \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}

amp; Now, P8+P6=18I−29Pamp; and P8−P6=8I−13Pamp;∴α=8,β=6,γ=18,δ=8amp;∴α+β+γ−δ=8+6+18−8=24\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}

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