Let P be a square matrix such that P^2=I-P.

Question

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

Accepted Answer

We have, ${P}^{2}=I-P$

$\begin{gathered} \Rightarrow {P}^{4} =(I-P{)}^{2}=I+{P}^{2}-2P \ \ =2I-3P $

$ \Rightarrow {P}^{8} =(2I-3P{)}^{2} \ \ =4I+9{P}^{2}-12P \ \ =13I-21P $

and

$ {P}^{6} =(I-P)(2I-3P) \ \ =2I-3P-2P+3{P}^{2} \ \ =5I-8P \end{gathered}$

$\begin{gathered} {P}^{8}+{P}^{6}=18I-29P \ {P}^{8}-{P}^{6}=8I-13P \ 	herefore \alpha =8,\beta =6,\gamma =18,\delta =8 \ 	herefore \alpha +\beta +\gamma -\delta =8+6+18-8=24 \end{gathered}$

Solutions · 1