Determinants – Cramer’s rule – JEE Main 24 January 2025 Shift 1

Solution:

For the system to have infinitely many solutions, by Cramer’s Rule, we must have $\Delta = \Delta_x = \Delta_y = \Delta_z = 0$.

First, set $\Delta = 0$:
$$ \Delta = \begin{vmatrix} 2 & -1 & 1 \\ 5 & \lambda & 3 \\ 100 & -47 & \mu \end{vmatrix} = 0 $$
$$ 2(\lambda\mu + 141) + 1(5\mu – 300) + 1(-235 – 100\lambda) = 0 $$
$$ 2\lambda\mu + 282 + 5\mu – 300 – 235 – 100\lambda = 0 $$
$$ 2\lambda\mu – 100\lambda + 5\mu – 253 = 0 \quad \dots(1) $$

Now, set $\Delta_x = 0$ (replace column 1 with constants 4, 12, 212):
$$ \Delta_x = \begin{vmatrix} 4 & -1 & 1 \\ 12 & \lambda & 3 \\ 212 & -47 & \mu \end{vmatrix} = 0 $$
Take 4 common from $C_1$:
$$ 4 \begin{vmatrix} 1 & -1 & 1 \\ 3 & \lambda & 3 \\ 53 & -47 & \mu \end{vmatrix} = 0 $$
$$ 1(\lambda\mu + 141) + 1(3\mu – 159) + 1(-141 – 53\lambda) = 0 $$
$$ \lambda\mu + 141 + 3\mu – 159 – 141 – 53\lambda = 0 $$
$$ \lambda\mu – 53\lambda + 3\mu – 159 = 0 $$
$$ \lambda(\mu – 53) + 3(\mu – 53) = 0 $$
$$ (\lambda + 3)(\mu – 53) = 0 $$
This gives two cases: $\lambda = -3$ or $\mu = 53$.

Check Case 1: $\lambda = -3$.
If we compute $\Delta_y$ with $\lambda = -3$, it does not vanish (leads to no solution). Thus, we discard $\lambda = -3$.

Check Case 2: $\mu = 53$.
Substitute $\mu = 53$ into equation (1):
$$ 2\lambda(53) – 100\lambda + 5(53) – 253 = 0 $$
$$ 106\lambda – 100\lambda + 265 – 253 = 0 $$
$$ 6\lambda + 12 = 0 $$
$$ \lambda = -2 $$

So, $\lambda = -2$ and $\mu = 53$.
We need the value of $\mu – 2\lambda$:
$$ \mu – 2\lambda = 53 – 2(-2) $$
$$ = 53 + 4 = 57 $$

Ans. (4)