JEE Main 2025 — 23 January, Evening Shift — Question 3

For the system to have no solution, the coefficient matrix must be singular (determinant zero) and at least one of the augmented matrix determinants must be non-zero.

First, find the determinant of the coefficient matrix:

$$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & \lambda \end{vmatrix}$$

Expanding along the first row:

$$D = 1\cdot(2\lambda - 25) - 1\cdot(\lambda - 5) + 1\cdot(5 - 2)$$ $$D = 2\lambda - 25 - \lambda + 5 + 3$$ $$D = \lambda - 17$$

For no solution, $D = 0$ ⇒ $\lambda = 17$.

Now check the determinant $D_z$ (with z-column replaced by constants):

$$D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 9 \\ 1 & 5 & \mu \end{vmatrix}$$

Expanding along the first row:

$$D_z = 1\cdot(2\mu - 45) - 1\cdot(\mu - 9) + 6\cdot(5 - 2)$$ $$D_z = 2\mu - 45 - \mu + 9 + 18$$ $$D_z = \mu - 18$$

For no solution, $D_z \neq 0$ ⇒ $\mu \neq 18$.

Thus, the system has no solution when $\lambda = 17$ and $\mu \neq 18$.