Question

A square matrix P satisfies \(P^2 = I - P\). If \(P^n = 5I - 8P\) then n is equal to

• A. 4
• B. 5
• C. 6
• D. 7

Hint:

Notice the sequence of coefficients of \(I\) and \(P\): \((0, 1) \rightarrow (1, -1) \rightarrow (-1, 2) \rightarrow (2, -3) \rightarrow (-3, 5) \rightarrow (5, -8)\). These are alternating signs of Fibonacci numbers!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are given a recurrence relation for the powers of a matrix \(P\): \(P^2 = I - P\). We need to find the power \(n\) such that \(P^n = 5I - 8P\). We will compute successive powers of \(P\) by substituting \(P^2\) with \(I - P\) to reduce higher powers to a linear combination of \(I\) and \(P\).


Step 2: Key Formula or Approach:
Multiply by \(P\) repeatedly and apply the substitution \(P^2 = I - P\) at each step.


Step 3: Detailed Explanation:
Given: \(P^2 = I - P\)
Multiply both sides by \(P\) to find \(P^3\): \[ P^3 = P(I - P) = P - P^2 \] Substitute \(P^2 = I - P\): \[ P^3 = P - (I - P) = 2P - I \] Multiply by \(P\) to find \(P^4\): \[ P^4 = P(2P - I) = 2P^2 - P \] Substitute \(P^2 = I - P\): \[ P^4 = 2(I - P) - P = 2I - 2P - P = 2I - 3P \] Multiply by \(P\) to find \(P^5\): \[ P^5 = P(2I - 3P) = 2P - 3P^2 \] Substitute \(P^2 = I - P\): \[ P^5 = 2P - 3(I - P) = 2P - 3I + 3P = 5P - 3I = -3I + 5P \] Multiply by \(P\) to find \(P^6\): \[ P^6 = P(-3I + 5P) = -3P + 5P^2 \] Substitute \(P^2 = I - P\): \[ P^6 = -3P + 5(I - P) = -3P + 5I - 5P = 5I - 8P \] We have reached the expression given in the question: \(P^6 = 5I - 8P\).
Therefore, comparing with \(P^n = 5I - 8P\), we get \(n = 6\).


Step 4: Final Answer:
The value of n is 6.