Matrices & Determinants
Powers of Matrices
Grade 12
Question:
<p>Let \(M\) denote the matrix \(\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\), where \(i^2 = -1\), and let \(I\) denote the identity matrix \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\). Then the matrix \(I + M + M^2 + M^3 + M^4 + \ldots + M^{2010}\) is equal to:</p>
<p>(a) \(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\)</p>
<p>(b) \(\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}\)</p>
<p>(c) \(\begin{pmatrix} 1 & i \\ i & 1 \end{pmatrix}\)</p>
<p>(d) \(\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}\)</p>
Step-by-Step Solution
Key Concept: Recognize that M is a matrix with finite order (M^4 = I), so the infinite-like sum becomes a finite geometric series that telescopes into a simple pattern based on the periodicity of powers.
<p><strong>Step 1:</strong> Compute powers of M:</p><p>M = ⎛0 i⎞, M² = ⎛0 i⎞⎛0 i⎞ = ⎛-1 0⎞ = -I</p><p> ⎝i 0⎠ ⎝i 0⎠⎝i 0⎠ ⎝0 -1⎠</p><p><strong>Step 2:</strong> Find the pattern: M³ = M·M² = M(-I) = -M, and M⁴ = M²·M² = (-I)(-I) = I</p><p>The powers repeat with period 4: M, -I, -M, I, M, -I, -M, I, ...</p><p><strong>Step 3:</strong> Group the sum by cycles of 4:</p><p>I + M + M² + M³ + M⁴ + ... + M²⁰¹⁰ = (I + M - I - M) + (I + M - I - M) + ... + (I + M - I - M) + (I + M)</p><p>Since 2011 = 4(502) + 3, we have 502 complete cycles (each summing to 0) plus the remainder terms.</p><p><strong>Step 4:</strong> The remainder terms are I + M + M² = I + M - I = ⎛0 i⎞</p><p> ⎝i 0⎠</p><p>∴ Answer: C</p>
Correct Answer: C