Matrices & Determinants
Series and Matrix Functions
Grade 12
Question:
<p>If <m>A = \begin{pmatrix} 2 & 1 \\ -4 & -2 \end{pmatrix}</m>, then <m>I + 2A + 3A^2 + \cdots + \infty</m> is</p>
<p>(a) <m>\begin{pmatrix} 4 & 1 \\ -4 & 0 \end{pmatrix}</m></p>
<p>(b) <m>\begin{pmatrix} 3 & 1 \\ -4 & -2 \end{pmatrix}</m></p>
<p>(c) <m>\begin{pmatrix} 5 & 2 \\ -3 & -8 \end{pmatrix}</m></p>
<p>(d) <m>\begin{pmatrix} 5 & 2 \\ -8 & -3 \end{pmatrix}</m></p>
Step-by-Step Solution
Key Concept: Recognize the infinite series as the derivative of a geometric series, and apply the formula for (I-A)⁻² where applicable.
<p><strong>Step 1:</strong> Recognize the series <m>S = I + 2A + 3A^2 + 4A^3 + \cdots</m> as <m>\sum_{n=0}^{\infty} (n+1)A^n</m>.</p><p><strong>Step 2:</strong> This is the derivative of <m>\sum_{n=0}^{\infty} A^n = (I-A)^{-1}</m> with respect to A, giving <m>S = (I-A)^{-2}</m>.</p><p><strong>Step 3:</strong> Calculate <m>I - A = \begin{pmatrix} -1 & -1 \\ 4 & 3 \end{pmatrix}</m>.</p><p><strong>Step 4:</strong> Find <m>(I-A)^{-1}</m> and then square it to get <m>(I-A)^{-2}</m>.</p><p>∴ Answer is (b).</p>
Correct Answer: B