Matrices & Determinants
System of Linear Equations
Grade 12
Question:
<p>Consider a system of linear equations \(3x + y - z = 0\), \(x - \dfrac{py}{4} + z = 2\) and \(2x - y + 2z = q\) where \(p, q \in I\) and \(p, q \in [1, 10]\), then identify the correct statement(s).</p><table border='1'><tr><th>List-I</th><th>List-II</th></tr><tr><td>(P) Number of ordered pairs \((p,q)\) for which system of equation has unique solution is</td><td>(1) 1</td></tr><tr><td>(Q) Number of ordered pairs \((p,q)\) for which system of equation has no solution is</td><td>(2) 9</td></tr><tr><td>(R) Number of ordered pairs \((p,q)\) for which system of equation has infinite solutions is</td><td>(3) 10</td></tr><tr><td>(S) Number of ordered pairs \((p,q)\) for which system of equation has atleast one solution is</td><td>(4) 90</td></tr><tr><td></td><td>(5) 91</td></tr></table>
<p>(a) P → 4; Q → 2; R → 1; S → 5</p>
<p>(b) P → 4; Q → 3; R → 1; S → 4</p>
<p>(c) P → 4; Q → 3; R → 2; S → 5</p>
<p>(d) P → 4; Q → 2; R → 2; S → 4</p>
Step-by-Step Solution
Key Concept: A system has a unique solution when det(A) ≠ 0, no solution when det(A) = 0 but rank(A) ≠ rank(A|B), and infinite solutions when det(A) = 0 and rank(A) = rank(A|B). Calculate the determinant and use consistency conditions to classify all 100 pairs (p,q).
\textbf{Step 1:} Write the coefficient matrix and find the determinant:
$$A = \begin{bmatrix} 3 & 1 & -1 \\ 1 & -\frac{p}{4} & 1 \\ 2 & -1 & 2 \end{bmatrix}$$
$$\det(A) = 3\left(-\frac{p}{2} + 1\right) - 1(2 - 2) - 1\left(-1 + \frac{p}{2}\right)$$
$$= 3\left(1 - \frac{p}{2}\right) - 0 - \left(\frac{p}{2} - 1\right)$$
$$= 3 - \frac{3p}{2} - \frac{p}{2} + 1 = 4 - 2p$$
\textbf{Step 2:} For unique solution: $\det(A) \neq 0 \implies p \neq 2$
Number of pairs $= 9 \times 10 = 90$ (all $(p,q)$ pairs except when $p = 2$)
\textbf{Step 3:} When $p = 2$, $\det(A) = 0$. Check the augmented matrix rank.
The equations become: $3x + y - z = 0$, $x - \frac{y}{2} + z = 2$, $2x - y + 2z = q$
\textbf{Step 4:} When $p = 2$, Row 3 satisfies: Row 3 $= \frac{2}{3}\text{Row 1} + \text{Row 2}$
This gives: $2x - y + 2z = 2$
For consistency with the third equation:
- Infinite solutions: $q = 2$ (1 pair)
- No solution: $q \in \{1, 3, 4, 5, 6, 7, 8, 9, 10\}$ (9 pairs)
\textbf{Step 5:} Count for $p = 2$:
\begin{itemize}
\item Infinite solutions: $q = 2 \to 1$ pair
\item No solution: $q \neq 2 \to 9$ pairs
\end{itemize}
\textbf{Step 6:} Verification:
(P) Unique solution: 90 pairs $\to$ Statement (4) \checkmark
(Q) No solution: 9 pairs $\to$ Statement (2) \checkmark
(R) Infinite solutions: 1 pair $\to$ Statement (1) \checkmark
(S) At least one solution: $90 + 1 = 91$ pairs $\to$ Statement (5) \checkmark
\therefore Answer: $P \to 4$; $Q \to 2$; $R \to 1$; $S \to 5$
Correct Answer: A