System of Linear Equations
DAILY_CHALLENGE
Grade None
Question:
Let $\alpha$, $\beta$ and $\gamma$ be real numbers. Consider the following system of linear equations:
$x+2y+z=7$
$x+\alpha z=11$
$2x-3y+\beta z=\gamma$
Match each entry in List-I to the correct entries in List-II.
**List-I**
(P) If $\beta=\dfrac{1}{2}(7\alpha-3)$ and $\gamma=28$, then the system has
(Q) If $\beta=\dfrac{1}{2}(7\alpha-3)$ and $\gamma\neq28$, then the system has
(R) If $\beta\neq\dfrac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma\neq28$, then the system has
(S) If $\beta\neq\dfrac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has
**List-II**
(1) a unique solution
(2) no solution
(3) infinitely many solutions
(4) $x=11$, $y=-2$ and $z=0$ as a solution
(5) $x=-15$, $y=4$ and $z=0$ as a solution
(P)→(3) (Q)→(2) (R)→(1) (S)→(4)
(P)→(3) (Q)→(2) (R)→(5) (S)→(4)
(P)→(2) (Q)→(1) (R)→(4) (S)→(5)
(P)→(2) (Q)→(1) (R)→(1) (S)→(3)
Step-by-Step Solution
Key Concept: Determinant condition governs uniqueness; consistency of the dependent system checked via row reduction
$\det(A)=7\alpha-2\beta-3$. When $\beta=\frac{1}{2}(7\alpha-3)$: $\det=7\alpha-(7\alpha-3)-3=0$.
(P) $\det=0$, $\gamma=28$: from equations, subtracting eq1−eq2: $2y+(1-\alpha)z=-4$. With eq3 and consistency analysis gives $6=\gamma-22$, so $\gamma=28$ is exactly the consistency condition. Infinitely many solutions→(3).
(Q) $\det=0$, $\gamma\neq28$: inconsistent. No solution→(2).
(R) $\alpha=1$, $\beta\neq2$ (since $\frac{1}{2}(7\cdot1-3)=2$): $\det=4-2\beta\neq0$. Unique solution. From eq1−eq2 (with $\alpha=1$): $2y=−4\Rightarrow y=-2$, $x+z=11$. Eq3: $28+(\beta-2)z=\gamma\Rightarrow z=(\gamma-28)/(\beta-2)\neq0$ when $\gamma\neq28$. Unique solution→(1).
(S) $\alpha=1$, $\beta\neq2$, $\gamma=28$: unique solution with $z=0$, $x=11$, $y=-2$. Solution is $(11,-2,0)$→(4).
Answer: (P)→(3),(Q)→(2),(R)→(1),(S)→(4) → A.
Correct Answer: A