Let α ,β in R be such that the system of linear equations x + 2 y + z = 5 , 2 x + y + α z = 5 , 8 x + 4 y + β z = 18 has no solution. Then β/α is equal to :

Mathematics · Determinants

JEE Main 2026 — 2 April, Morning Shift — Question 22

Let $\alpha ,\beta \in \mathbf{R}$ be such that the system of linear equations $\mathbf{x} + 2\mathbf{y} + \mathbf{z} = 5$ , $2\mathbf{x} + \mathbf{y} + \alpha \mathbf{z} = 5$ , $8\mathbf{x} + 4\mathbf{y} + \beta \mathbf{z} = 18$ has no solution. Then $\frac{\beta}{\alpha}$ is equal to :

Option A:
-4
Option B:
4
Correct
Option C:
8
Option D:
-8

Answer: B

Step-by-step solution

$x+2 y+z=5$ $2 x+y+\alpha z=5$ $8 x+4 y+\beta z=18$ for no solution

\begin{aligned} & \Delta=\left|\begin{array}{lll} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{array}\right|=0 \\& \Delta=1(\beta-4 \alpha)-2(2 \beta-8 \alpha)+1(0)=0 \\& =\beta-4 \alpha-4 \beta+16 \alpha=0 \\& \Rightarrow 4 \alpha=\beta \end{aligned}

$\frac{\beta}{\alpha}=4$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 2 April, Morning Shift
Subject: Mathematics
Chapter: Determinants
Topic: System of Linear Equations using Determinants