JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 5)
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 :
18
22
24
40
Explanation
We have, $P^2=I-P$
$$ \begin{aligned} \Rightarrow P^4 & =(I-P)^2=I+P^2-2 P \\\\ & =2 I-3 P \text { [Using Eq. (i)] } \end{aligned} $$
$$ \begin{aligned} \Rightarrow P^8 & =(2 I-3 P)^2 \\\\ & =4 I+9 P^2-12 P \\\\ & =13 I-21 P \text { [Using Eq. (i)] } \end{aligned} $$
and
$$ \begin{aligned} P^6 & =(I-P)(2 I-3 P) \\\\ & =2 I-3 P-2 P+3 P^2 \\\\ & =5 I-8 P \text { [Using Eq. (i)] } \end{aligned} $$
$\begin{aligned} & \text { Now, } P^8+P^6=18 I-29 P \\\\ & \text { and } P^8-P^6=8 I-13 P \\\\ & \therefore \alpha=8, \beta=6, \gamma=18, \delta=8 \\\\ & \therefore \alpha+\beta+\gamma-\delta=8+6+18-8=24\end{aligned}$
$$ \begin{aligned} \Rightarrow P^4 & =(I-P)^2=I+P^2-2 P \\\\ & =2 I-3 P \text { [Using Eq. (i)] } \end{aligned} $$
$$ \begin{aligned} \Rightarrow P^8 & =(2 I-3 P)^2 \\\\ & =4 I+9 P^2-12 P \\\\ & =13 I-21 P \text { [Using Eq. (i)] } \end{aligned} $$
and
$$ \begin{aligned} P^6 & =(I-P)(2 I-3 P) \\\\ & =2 I-3 P-2 P+3 P^2 \\\\ & =5 I-8 P \text { [Using Eq. (i)] } \end{aligned} $$
$\begin{aligned} & \text { Now, } P^8+P^6=18 I-29 P \\\\ & \text { and } P^8-P^6=8 I-13 P \\\\ & \therefore \alpha=8, \beta=6, \gamma=18, \delta=8 \\\\ & \therefore \alpha+\beta+\gamma-\delta=8+6+18-8=24\end{aligned}$
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