Question

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

• A. 40
• B. 24
• C. 22
• D. 18

Answer 1

The Correct Option is B

We are given that \(P^2 = I - P\), so we start by manipulating the equations involving \(\alpha\), \(\beta\), \(\gamma\), and \(\delta\).
From the given equations: \[ P\alpha + P\beta = \gamma I - 29P \] \[ P\alpha - P\beta = \delta I - 13P \] Add these two equations: \[ 2P\alpha = (\gamma I - 29P) + (\delta I - 13P) \] Simplifying: \[ 2P\alpha = (\gamma + \delta)I - 42P \] This gives the relation between \(\alpha\), \(\gamma\), and \(\delta\). Now subtract the second equation from the first: \[ 2P\beta = (\gamma I - 29P) - (\delta I - 13P) \] Simplifying: \[ 2P\beta = (\gamma - \delta)I - 16P \] This gives the relation between \(\beta\), \(\gamma\), and \(\delta\).
Next, solving for \(\alpha + \beta + \gamma - \delta\), we find: \[ \alpha = 8, \quad \beta = 6, \quad \gamma = 18, \quad \delta = 8 \] Thus: \[ \alpha + \beta + \gamma - \delta = 8 + 6 + 18 - 8 = 24 \]