Matrices & DeterminantsmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Inconsistent Linear System: β/α = 4 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
Let α,β∈R\alpha,\beta\in\mathbb{R} be such that the system of linear equations
x+2y+z=5,2x+y+αz=5,8x+4y+βz=18x+2y+z=5,\quad 2x+y+\alpha z=5,\quad 8x+4y+\beta z=18
 has no solution. Then βα\dfrac{\beta}{\alpha} is equal to
A−4-4
B44correct
C88
D−8-8
Solution
Step 1: For the system to have no solution (a necessary condition here), the coefficient determinant must be zero:
Δ=∣12121α84β∣=0.\Delta=\begin{vmatrix}1&2&1\\2&1&\alpha\\8&4&\beta\end{vmatrix}=0.
 Step 2: Expand along the first row:
1(β−4α)−2(2β−8α)+1(8−8)=0.1(\beta-4\alpha)-2(2\beta-8\alpha)+1(8-8)=0.
 Step 3: Simplify:
β−4α−4β+16α=0 ⇒ −3β+12α=0 ⇒ 4α=β.\beta-4\alpha-4\beta+16\alpha=0\ \Rightarrow\ -3\beta+12\alpha=0\ \Rightarrow\ 4\alpha=\beta.
 Step 4: Therefore
βα=4.\frac{\beta}{\alpha}=4.
 Correct answer: (2)
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