Matrices & Determinants
Summation of Determinants
Grade 12
Question:
<p>If \[\Delta_r = \begin{vmatrix} r & 612 & 915 \\ 101r^2 & 2r & 3r \\ r & \dfrac{1}{r} & \dfrac{1}{r^2} \end{vmatrix}\], then the value of \(\displaystyle\lim_{n \to \infty} \dfrac{1}{n^3} \sum_{r=1}^{n} \Delta_r\) is ________.</p>
Step-by-Step Solution
Key Concept: Factor out r from rows to reveal that Δ_r = r³·D where D is independent of r, then use the summation formula Σr³ = [n(n+1)/2]² to evaluate the limit as a Riemann sum.
<p><strong>Step 1: Factor the determinant</strong></p><p>From Row 1: factor out r from columns 2,3 (gives r²)</p><p>From Row 2: factor out r from all columns (gives r³)</p><p>From Row 3: factor out 1/r from all columns (gives 1/r³)</p><p>Net result: Δ_r = r³ · D, where</p><p>$$D = \begin{vmatrix} 1 & 612 & 915 \\ 101 & 2 & 3 \\ 1 & 1 & 1 \end{vmatrix}$$</p><p><strong>Step 2: Compute D using row operations</strong></p><p>R₂ → R₂ - 101R₁; R₃ → R₃ - R₁:</p><p>$$D = \begin{vmatrix} 1 & 612 & 915 \\ 0 & -61810 & -92684 \\ 0 & -611 & -914 \end{vmatrix}$$</p><p>Expanding along column 1:</p><p>$$D = (-61810)(-914) - (-92684)(-611) = 56535140 - 56606324 = -71184$$</p><p><strong>Step 3: Evaluate the limit</strong></p><p>$$\lim_{n \to \infty} \frac{1}{n^3} \sum_{r=1}^{n} \Delta_r = \lim_{n \to \infty} \frac{D}{n^3} \sum_{r=1}^{n} r^3 = D · \lim_{n \to \infty} \frac{1}{n^3} · \frac{n^2(n+1)^2}{4}$$</p><p>$$= D · \lim_{n \to \infty} \frac{(n+1)^2}{4n} = D · \frac{1}{4} = -71184 · \frac{1}{4} = -17796$$</p><p>∴ Answer: <strong>-17796</strong></p>
Correct Answer: -17796