JEE Mathematics 2026 Question with Solution

Given: The system is

$$\begin{aligned} 3x + y + 4z &= 3 \\ 2x + \alpha y - z &= -3 \\ x + 2y + z &= 4 \end{aligned}$$

Find: The value of $\alpha$ for which the system has no solution.

For a system of three linear equations, a necessary condition for no solution is that the determinant of the coefficient matrix is zero, and at least one corresponding Cramer's determinant is non-zero.

So first calculate

$$\Delta = \begin{vmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix}$$

Expanding,

$$\Delta = 3(\alpha + 2) - 1(2 + 1) + 4(4 - \alpha)$$ $$\Delta = 3\alpha + 6 - 3 + 16 - 4\alpha = 19 - \alpha$$

For no solution, set

$$\Delta = 0$$

So,

$$19 - \alpha = 0 \implies \alpha = 19$$

Now verify inconsistency using

$$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 \end{vmatrix}$$

From the extracted solution,

$$\Delta_x = 3(19 + 2) - 1(-3 + 4) + 4(-6 - 76) = 63 - 1 - 328 \neq 0$$

Thus $\Delta = 0$ and $\Delta_x \neq 0$, so the system has no solution.

Therefore, the correct option is A, and the value of $\alpha$ is $19$.

Given: A parameter-dependent linear system in $x, y, z$.

Find: When the system becomes inconsistent.

A linear system has:

• a unique solution when the coefficient determinant is non-zero,
• infinitely many or no solution when the coefficient determinant is zero.

To distinguish between these last two cases, check whether at least one numerator determinant is non-zero. Here the solution shows that

$$\Delta = 19 - \alpha$$

Hence the critical value is obtained from

$$19 - \alpha = 0$$

which gives

$$\alpha = 19$$

Then the solution verifies that

$$\Delta_x \neq 0$$

Therefore the system is inconsistent, so it has no solution.

Hence, the required value is $19$.