JEE Main 2026 — Mathematics Algebra

The given system of linear equations is:

$$x + 2y + z = 5$$x+2y+z=5

$$2x + y + \alpha z = 5$$2x+y+αz=5

$$8x + 4y + \beta z = 18$$8x+4y+βz=18

For the system to have no solution, the determinant of the coefficient matrix $D$D must be zero.

$$D = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix} = 0$$D=​128​214​1αβ​​=0

Expanding along the first row:

$$1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(8 - 8) = 0$$1(β−4α)−2(2β−8α)+1(8−8)=0

$$\beta - 4\alpha - 4\beta + 16\alpha = 0$$β−4α−4β+16α=0

$$12\alpha - 3\beta = 0$$12α−3β=0

$$\beta = 4\alpha$$β=4α

To verify the no solution condition, substitute $\beta = 4\alpha$β=4α into the third equation:

$$8x + 4y + 4\alpha z = 18$$8x+4y+4αz=18

$$4(2x + y + \alpha z) = 18$$4(2x+y+αz)=18

$$2x + y + \alpha z = \dfrac{9}{2}$$2x+y+αz=29​

However, the second equation is $2x + y + \alpha z = 5$2x+y+αz=5. This results in a contradiction since $\dfrac{9}{2} \neq 5$29​=5. Thus, the system has no solution when $\beta = 4\alpha$β=4α.

Therefore, $\dfrac{\beta}{\alpha} = 4$αβ​=4.

Answer: $4$4