Matrices & Determinants
Powers of Matrices
Grade 12
Question:
<p>Matrix A such that <m>A^2 = 2A - I</m>, where I is the identity matrix. Then, for <m>n \geq 2</m>, <m>A^n</m> is equal to</p>
<p>(a) <m>nA - (n-1)I</m></p>
<p>(b) <m>nA - I</m></p>
<p>(c) <m>2^{n-1}A - (n-1)I</m></p>
<p>(d) <m>2^{n-1}A - I</m></p>
Step-by-Step Solution
Key Concept: Use the given recurrence relation A² = 2A - I and proof by mathematical induction to find the general form of Aⁿ.
<p><strong>Step 1:</strong> From <m>A^2 = 2A - I</m>, rewrite as <m>A^2 - 2A + I = 0</m>, or <m>(A-I)^2 = 0</m>.</p><p><strong>Step 2:</strong> Use induction. For <m>n=2</m>: <m>A^2 = 2A - I = 2A - (2-1)I</m> ✓</p><p><strong>Step 3:</strong> Assume <m>A^k = kA - (k-1)I</m>. Then <m>A^{k+1} = A \cdot A^k = A[kA - (k-1)I] = k(2A-I) - (k-1)A = (k+1)A - kI</m>.</p><p>∴ Answer is (a) <m>nA - (n-1)I</m>.</p>
Correct Answer: A