Matrices & Determinants
Matrix Powers and Series
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 periodic with period 4 (M⁴ = I), so the infinite series becomes a finite sum of complete cycles plus remainder terms. Sum the geometric series using the periodicity pattern to find the result.
<p><strong>Step 1:</strong> Compute powers of M.</p><p>M = <begin>pmatrix</begin> 0 & i \\ i & 0 \end{pmatrix}</p><p>M² = <begin>pmatrix</begin> 0 & i \\ i & 0 \end{pmatrix} <begin>pmatrix</begin> 0 & i \\ i & 0 \end{pmatrix} = <begin>pmatrix</begin> -1 & 0 \\ 0 & -1 \end{pmatrix} = -I</p><p>M³ = M² · M = -I · M = <begin>pmatrix</begin> 0 & -i \\ -i & 0 \end{pmatrix} = -M</p><p>M⁴ = M² · M² = (-I)(-I) = I</p><p><strong>Step 2:</strong> Identify the periodic pattern.</p><p>Since M⁴ = I, the powers repeat with period 4: M, M², M³, M⁴ = I, M⁵ = M, ...</p><p><strong>Step 3:</strong> Group terms in cycles of 4.</p><p>S = I + M + M² + M³ + M⁴ + M⁵ + ... + M²⁰¹⁰</p><p>Since 2010 = 4(502) + 2, we have 502 complete cycles plus 3 extra terms (I, M, M²).</p><p>One complete cycle sum: I + M + M² + M³ = I + M + (-I) + (-M) = 0</p><p><strong>Step 4:</strong> Calculate the final sum.</p><p>S = 502(0) + (I + M + M²) = I + M + (-I) = M = <begin>pmatrix</begin> 0 & i \\ i & 0 \end{pmatrix}</p><p>∴ Answer: C</p>
Correct Answer: C