JEE Main 2026 · 2 Apr · Shift 1 · Mathematics Q3

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MathematicsSingle correct+4 / -1Question 3 of 75

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

-4

4

8

-8

Answer and worked solution

Official answer: B

No solution needs $\Delta=0$ :
$\begin{vmatrix}1&2&1\\2&1&\alpha\\8&4&\beta\end{vmatrix}=(\beta-4\alpha)-2(2\beta-8\alpha)=12\alpha-3\beta=0\implies\beta=4\alpha.$

Inconsistency check. With $\beta=4\alpha$ the third left side is 4 times the second, $4(2x+y+\alpha z)$ , which would have to equal 20, not 18. So the system is inconsistent.

Ratio. $\dfrac\beta\alpha=4$ .

(B)

Topics:Systems of linear equations Determinants and properties

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JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 3