Algebra – Matrices and Determinants – JEE Main 21 Jan 2026 Shift 2

Solution:

For a system of linear equations to have no solution, the determinant of the coefficient matrix ($\Delta$) must be zero, and at least one of the determinants $\Delta_x, \Delta_y, \Delta_z$ must be non-zero.

$$\Delta = \begin{vmatrix} 3 & 1 & 4 \\ 2 & a & -1 \\ 1 & 2 & 1 \end{vmatrix} = 0$$

$$\therefore 3(a(1) – (-1)(2)) – 1(2(1) – (-1)(1)) + 4(2(2) – a(1)) = 0$$

$$\therefore 3(a + 2) – 1(2 + 1) + 4(4 – a) = 0$$

$$\therefore 3a + 6 – 3 + 16 – 4a = 0$$

$$\therefore -a + 19 = 0$$

$$\therefore a = 19$$

Now, we must verify that for $a=19$, at least one of $\Delta_x, \Delta_y, \Delta_z$ is non-zero. Let’s calculate $\Delta_x$:
$$\Delta_x = \begin{vmatrix} 3 & 1 & 4 \\ -3 & 19 & -1 \\ 4 & 2 & 1 \end{vmatrix}$$

$$\therefore \Delta_x = 3(19 – (-2)) – 1(-3 – (-4)) + 4(-6 – 76)$$

$$\therefore \Delta_x = 3(21) – 1(1) + 4(-82)$$

$$\therefore \Delta_x = 63 – 1 – 328 \neq 0$$

Since $\Delta = 0$ and $\Delta_x \neq 0$, the system has no solution for $a=19$.

Ans. (1)