Determinants – Cramer’s Rule – JEE Main – 23 January 2025 Shift 2

Solution:

For the system of equations to have no solution, the determinant of the coefficient matrix ($D$) must be zero, and at least one of the determinants $D_x, D_y, D_z$ must be non-zero.

The determinant $D$ is given by:
$$D = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 2 & 5 \\ 1 & 5 & \lambda \end{vmatrix}$$
$$D = 1(2\lambda – 25) – 1(\lambda – 5) + 1(5 – 2)$$
$$D = 2\lambda – 25 – \lambda + 5 + 3$$
$$D = \lambda – 17$$

For no solution (or infinite solutions), $D = 0 \Rightarrow \lambda – 17 = 0 \Rightarrow \lambda = 17$.

Now, calculate $D_z$ by replacing the third column with the constant terms:
$$D_z = \begin{vmatrix} 1 & 1 & 6 \\ 1 & 2 & 9 \\ 1 & 5 & \mu \end{vmatrix}$$
$$D_z = 1(2\mu – 45) – 1(\mu – 9) + 6(5 – 2)$$
$$D_z = 2\mu – 45 – \mu + 9 + 18$$
$$D_z = \mu – 18$$

For the system to have **no solution**, we need $D = 0$ and at least one of $D_x, D_y, D_z \ne 0$.
Using $D_z \ne 0$, we get:
$$\mu – 18 \ne 0 \Rightarrow \mu \ne 18$$

Thus, the condition is $\lambda = 17$ and $\mu \ne 18$.

Ans. (1)