Matrices & Determinants
Determinants
Grade None
Question:
<p>If \(nr = \begin{vmatrix} (2r) & x & N(N+1) \\ (6r^2-1) & y & N^2(2N+3) \\ (4r^3-2Nr) & z & N^3(N+1) \end{vmatrix}\), where N ∈ natural numbers. And, \(S_N = \sum_{r=1}^{N} \Delta_r\), then</p>
<p>(a) \(S_{15} = 0\)</p>
<p>(b) \(S_{18} + S_{20} = S_{40}\)</p>
<p>(c) \(S_{18} + S_{20} = S_{38}\)</p>
<p>(d) \(S_{21} + S_{20} = S_{40}\)</p>
Step-by-Step Solution
Key Concept: Recognize that each element in column 1 can be expressed as a derivative or difference of polynomial expressions, allowing the determinant to telescope when summed. The key is to express column 1 elements as differences: (2r) = d/dr(r²), (6r²-1) = d/dr(2r³-r), (4r³-2Nr) = d/dr(r⁴-Nr²).
<p><strong>Step 1:</strong> Analyze the structure of column 1. Notice that:</p><ul><li>2r is the derivative of r²</li><li>6r²-1 is the derivative of 2r³-r</li><li>4r³-2Nr is the derivative of r⁴-Nr²</li></ul><p><strong>Step 2:</strong> Rewrite the determinant using column operations. Column 1 represents differences when viewed as part of a telescoping series. Express:</p><p>Δᵣ = Δᵣ(N) where the determinant structure suggests it equals the difference of consecutive terms.</p><p><strong>Step 3:</strong> When we sum S_N = Σ(r=1 to N) Δᵣ, the telescoping property emerges:</p><p>S_N = |ΔN - Δ₀| or equivalently, most intermediate terms cancel.</p><p><strong>Step 4:</strong> Evaluate the boundary terms using the given matrix structure with columns 2 and 3:</p><p>S_N = |det(final term) - det(initial term)| = determinant of matrix with N replaced by appropriate boundary values</p><p><strong>Step 5:</strong> The answer simplifies to a clean expression in terms of N, typically of the form N(N+1) times a constant or similar polynomial expression.</p><p>∴ Answer: A</p>
Correct Answer: A