Q35 / 75 · JEE Main 2026 (21 January Shift 2) · Mathematics
If the system of equations
$3 x+y+4 z=3$
$2 x+\alpha y-z=-3$
$x+2 y+z=4$
has no solution, then the value of $\alpha$ is equal to :
$3 x+y+4 z=3$
$2 x+\alpha y-z=-3$
$x+2 y+z=4$
has no solution, then the value of $\alpha$ is equal to :
- Correct answerA$19$
- B4
- C13
- D23
Solution
For no solution $\Delta = 0$
$\begin{vmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$
$\Rightarrow 3(\alpha + 2) + 1(-1 - 2) + 4(4 - \alpha) = 0$
$\Rightarrow 19 - \alpha = 0 \Rightarrow \boxed{\alpha = 19}$
& for $\alpha = 19$
$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 \end{vmatrix} = 3(21) + 1(-1) + 4(-82)$
$\neq 0$
$\therefore$ no solution for $\alpha = 19$
$\begin{vmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$
$\Rightarrow 3(\alpha + 2) + 1(-1 - 2) + 4(4 - \alpha) = 0$
$\Rightarrow 19 - \alpha = 0 \Rightarrow \boxed{\alpha = 19}$
& for $\alpha = 19$
$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 \end{vmatrix} = 3(21) + 1(-1) + 4(-82)$
$\neq 0$
$\therefore$ no solution for $\alpha = 19$
Hence, the correct answer is option: 'A'.