Determinants – Cramer’s rule – JEE Main 24 January 2025 Shift 2

Solution:

For the system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix ($D$) and the determinants formed by replacing columns with the constant terms ($D_1, D_2, D_3$) must all be zero.

First, we find $D$:
$$ D = \begin{vmatrix} 1 & 2 & -3 \\ 2 & \lambda & 5 \\ 14 & 3 & \mu \end{vmatrix} = 0 $$
$$ 1(\lambda \mu – 15) – 2(2\mu – 70) – 3(6 – 14\lambda) = 0 $$
$$ \lambda \mu – 15 – 4\mu + 140 – 18 + 42\lambda = 0 $$
$$ \lambda \mu + 42\lambda – 4\mu + 107 = 0 \quad \dots(1) $$

Now consider $D_2$ (replacing the y-column):
$$ D_2 = \begin{vmatrix} 1 & 2 & -3 \\ 2 & 5 & 5 \\ 14 & 33 & \mu \end{vmatrix} = 0 $$
Expanding along the first row:
$$ 1(5\mu – 165) – 2(2\mu – 70) – 3(66 – 70) = 0 $$
$$ 5\mu – 165 – 4\mu + 140 + 12 = 0 $$
$$ \mu – 13 = 0 \Rightarrow \mu = 13 $$

Now consider $D_3$ (replacing the z-column):
$$ D_3 = \begin{vmatrix} 1 & 2 & 2 \\ 2 & \lambda & 5 \\ 14 & 3 & 33 \end{vmatrix} = 0 $$
Applying $R_3 \to R_3 – 14R_1$ and $R_2 \to R_2 – 2R_1$:
$$ \begin{vmatrix} 1 & 2 & 2 \\ 0 & \lambda-4 & 1 \\ 0 & -25 & 5 \end{vmatrix} = 0 $$
$$ 1(5(\lambda-4) – (-25)) = 0 $$
$$ 5\lambda – 20 + 25 = 0 $$
$$ 5\lambda + 5 = 0 \Rightarrow \lambda = -1 $$

Checking in equation (1): $(-1)(13) + 42(-1) – 4(13) + 107 = -13 – 42 – 52 + 107 = -107 + 107 = 0$. (Satisfied).

We need to find $\lambda + \mu$:
$$ \lambda + \mu = -1 + 13 = 12 $$

Ans. (4)