JEE Main 2026 — Determinants Question with Solution

The given system of linear equations is:
$x+2y+z=5$
$2x+y+\alpha z=5$
$8x+4y+\beta z=18$

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

$D=\left|\begin{array}{ccc}1& 2& 1\\ 2& 1& \alpha \\ 8& 4& \beta \end{array}\right|=0$

Expanding along the first row:

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

$\beta -4\alpha -4\beta +16\alpha =0$

$12\alpha -3\beta =0$

$\beta =4\alpha$

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

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

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

$2x+y+\alpha z=\frac{9}{2}$

However, the second equation is $2x+y+\alpha z=5$. This results in a contradiction since $\frac{9}{2}\ne 5$. Thus, the system has no solution when $\beta =4\alpha$.

Therefore, $\frac{\beta }{\alpha }=4$.

Answer: $4$