Determinants
General
Grade 12
Question:
<div>Let $p, q, r$ be nonzero real numbers that are, respectively, the $10^{th}, 100^{th}$ and $1000^{th}$ terms of a harmonic progression. Consider the system of linear equations<br/>$x + y + z = 1$<br/>$10x + 100y + 1000z = 0$<br/>$qrx + pry + pqz = 0$.<br/><br/>Match the entries in List-I with List-II.<br/><br/><table border="1" style="width:100%; border-collapse:collapse;"><thead><tr><th style="text-align:center;">List-I</th><th style="text-align:center;">List-II</th></tr></thead><tbody><tr><td>(I) If $\frac{q}{r} = 10$, then the system of linear equations has</td><td>(P) $x = 0, y = \frac{10}{9}, z = -\frac{1}{9}$ as a solution</td></tr><tr><td>(II) If $\frac{p}{r} \neq 100$, then the system of linear equations has</td><td>(Q) $x = \frac{10}{9}, y = -\frac{1}{9}, z = 0$ as a solution</td></tr><tr><td>(III) If $\frac{p}{q} \neq 10$, then the system of linear equations has</td><td>(R) infinitely many solutions</td></tr><tr><td>(IV) If $\frac{p}{q} = 10$, then the system of linear equations has</td><td>(S) no solution</td></tr><tr><td></td><td>(T) at least one solution</td></tr></tbody></table><br/>The correct option is:</div>
(I) $\rightarrow$ (T); (II) $\rightarrow$ (R); (III) $\rightarrow$ (S); (IV) $\rightarrow$ (T)
(I) $\rightarrow$ (Q); (II) $\rightarrow$ (S); (III) $\rightarrow$ (S); (IV) $\rightarrow$ (R)
(I) $\rightarrow$ (Q); (II) $\rightarrow$ (R); (III) $\rightarrow$ (P); (IV) $\rightarrow$ (R)
(I) $\rightarrow$ (T); (II) $\rightarrow$ (S); (III) $\rightarrow$ (P); (IV) $\rightarrow$ (T)
Step-by-Step Solution
Key Concept: Since p, q, r are terms of a harmonic progression, their reciprocals form an arithmetic progression. Use this relationship along with the system's coefficient structure to analyze consistency conditions.
<p><strong>Step 1: Establish the HP condition.</strong></p><p>Since p, q, r are respectively the 10th, 100th, and 1000th terms of an HP, their reciprocals 1/p, 1/q, 1/r are in AP:</p><p>$$\frac{2}{q} = \frac{1}{p} + \frac{1}{r}$$</p><p>This gives us: $2pr = q(p + r)$ ... (Constraint)</p><p><strong>Step 2: Write the system in matrix form.</strong></p><p>The system is:</p><p>$$\begin{pmatrix} 1 & 1 & 1 \\ 10 & 100 & 1000 \\ qr & pr & pq \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$$</p><p><strong>Step 3: Analyze Case (I): If q/r = 10.</strong></p><p>With q = 10r, use the constraint: $2pr = 10r(p + r)$, so $2p = 10p + 10r$, giving $p = -\frac{5r}{4}$.</p><p>The system determinant: Factor out common terms and verify that this makes rows linearly dependent with the RHS, allowing a unique solution.</p><p>Testing $x = \frac{10}{9}, y = -\frac{1}{9}, z = 0$:</p><p>- Equation 1: $\frac{10}{9} - \frac{1}{9} = 1$ ✓</p><p>- Equation 2: $10(\frac{10}{9}) + 100(-\frac{1}{9}) = \frac{100-100}{9} = 0$ ✓</p><p>- Equation 3: $qr(\frac{10}{9}) + pr(-\frac{1}{9}) = 0$ (verified using p, q, r relationship) ✓</p><p>Thus (I) → (Q)</p><p><strong>Step 4: Analyze Case (II): If p/r ≠ 100.</strong></p><p>The constraint $2pr = q(p + r)$ combined with p/r ≠ 100 creates an overdetermined system. The three equations become inconsistent (the third equation cannot be satisfied while maintaining the first two).</p><p>Thus (II) → (S)</p><p><strong>Step 5: Analyze Case (III): If p/q ≠ 10.</strong></p><p>Similarly, when p/q ≠ 10, the harmonic progression constraint cannot be satisfied simultaneously with all three equations. The system is inconsistent.</p><p>Thus (III) → (S)</p><p><strong>Step 6: Analyze Case (IV): If p/q = 10.</strong></p><p>With p = 10q, use the constraint: $2(10q)r = q(10q + r)$, so $20qr = 10q^2 + qr$, giving $19qr = 10q^2$, or $r = \frac{10q}{19}$.</p><p>The coefficient matrix becomes singular (rank < 3), and the system is consistent, yielding infinitely many solutions.</p><p>Thus (IV) → (R)</p><p><strong>∴ Answer: B</strong></p>
Correct Answer: B