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

To solve this problem, we are given a square matrix \( P \) that satisfies the equation \( P^2 = I - P \). Our task is to find \( n \) such that \( P^n = 5I - 8P \).

Firstly, let's explore the properties of the matrix \( P \) using the given equation:

1. Start with the equation: \( P^2 = I - P \).
   Rearranging gives \( P^2 + P - I = 0 \). This is a quadratic equation in terms of matrix \( P \).
2. In matrix algebra, if a matrix satisfies a polynomial equation, we can often find powers of the matrix by substituting back. In this case, we will use this equation to express higher powers of \( P \).
   We need to find \( P^n = 5I - 8P \).
3. Firstly, try to build a pattern by calculating subsequent powers of \( P \):
   - Using \( P^2 = I - P \), calculate \( P^3 \): \(P^3 = P \cdot P^2 = P(I - P) = P - P^2 = P - (I - P) = 2P - I\)
4. Now, move to \( P^4 \):
   \(P^4 = P \cdot P^3 = P \cdot (2P - I) = 2P^2 - P = 2(I - P) - P = 2I - 2P - P = 2I - 3P\)
5. We continue to find \( P^5 \):
   \(P^5 = P \cdot P^4 = P \cdot (2I - 3P) = 2P - 3P^2 = 2P - 3(I - P) = 5P - 3I\)
6. Finally, calculate \( P^6 \):
   \(P^6 = P \cdot P^5 = P \cdot (5P - 3I) = 5P^2 - 3P = 5(I - P) - 3P = 5I - 8P\)
7. Thus, we see that \( P^6 = 5I - 8P \). Therefore, \( n = 6 \).

The correct answer is 6. This matches the condition given in the problem statement.