Question

If an n x n matrix A satisfies the equation \(A^2 - 2A + I = 0\), then:
1. \(A^3 = 3A - 2I, A^4 = 4A - 3I\)
2. \(A^3 = 2A - I, A^4 = 4A - 3I\)
3. \(A^3 = 2A^2 - 3I, A^4 = 3A^2 - 4I\)
4. \(A^3 = 4A - 3I, A^4 = 3A - 2I\)

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Exam Tip:
When a matrix satisfies a polynomial equation, we can use it to reduce higher powers:

• Use the relation \(A^2 = 2A - I\).
• Multiply repeatedly to find \(A^3, A^4, \dots\).
• This is similar to reducing powers in modular arithmetic.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given a matrix equation \(A^2 - 2A + I = 0\), which can be rewritten as \(A^2 = 2A - I\). We need to find expressions for higher powers of A.

Step 2: Key Formula or Approach:
We can use the given relation to reduce \(A^3\) and \(A^4\) to linear combinations of A and I.

Step 3: Detailed Explanation:
Given \(A^2 = 2A - I\).
• Find \(A^3\):
\(A^3 = A^2 \cdot A = (2A - I)A = 2A^2 - A = 2(2A - I) - A = 4A - 2I - A = 3A - 2I\).
• Find \(A^4\):
\(A^4 = A^3 \cdot A = (3A - 2I)A = 3A^2 - 2A = 3(2A - I) - 2A = 6A - 3I - 2A = 4A - 3I\). So, \(A^3 = 3A - 2I\) and \(A^4 = 4A - 3I\).
This matches option 1.

Step 4: Final Answer:
Therefore, option (A) is correct.