An ordered pair (α, β) for which the system of linear equations (1 + α)x + + z = 2 + (1 + β)y + z = 3 + + 2z = 2 has a unique solution, is :

Correct answer: C. (2, 4) For unique solution | cases 1 + α & β & 1 α & 1 + β & 1 α & β & 2 cases | ≠ 0 Applying C _ 1 → C _ 1 + C _ 2 + C _ 3 , we get | cases α + β + 2 & β & 1 α + β + 2 & 1 + β & 1 α + β + 2 & β & 2 cases | ≠ 0 ⇒ (α + β + 2) | cases 1 & β & 1 1 & 1 + β & 1 1 & β & 2 cases | ≠ 0 ⇒ α + β + 2 ≠ 0 Clearly, point (2, 4) satisfying the given condition.

JEE Main 201912 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 12 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Determinants

JEE Main 2019 · 12 January · Morning Shift · Mathematics · Determinants

An ordered pair (α, β) for which the system of linear equations (1...

An ordered pair

\left(\alpha, \beta\right)

for which the system of linear equations
\left(1 + \alpha\right)x + \betay + z = 2

<br />

\alphax + \left(1 + \beta\right)y + z = 3

<br />

\alphax + \betay + 2z = 2$ has a unique solution, is :

Options

A $\left(- 3, 1\right)$
B $\left(1, - 3\right)$
C $\left(2, 4\right)$
D $\left(- 4, 2\right)$

Correct answer

C. $\left(2, 4\right)$

Step-by-step solution

For unique solution $\left|$ \begin{cases}1 + \alpha & \beta & 1 \\ \alpha & 1 + \beta & 1 \\ \alpha & \beta & 2\end{cases} $\right| \ne 0$

Applying ${C}_{1} \to {C}_{1} + {C}_{2} + {C}_{3}$ , we get

\left| $\begin{cases}\alpha + \beta + 2 & \beta & 1 \\ \alpha + \beta + 2 & 1 + \beta & 1 \\ \alpha + \beta + 2 & \beta & 2\end{cases}$ \right| \ne 0 $$ ⇒ $\left(\alpha + \beta + 2\right)$ \left| $\begin{cases}1 & \beta & 1 \\ 1 & 1 + \beta & 1 \\ 1 & \beta & 2\end{cases}$ \right| \ne 0 $$ ⇒ \alpha + \beta + 2 \ne 0 $Clearly, point \left(2, 4\right)$ satisfying the given condition.

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