JEE Main 2025 Jan 24 Shift 2, Mathematics Q11: Linear Equations And Cramer’s Rule

For infinitely many solutions, all of $D$, $D_1$, $D_2$, $D_3$ must equal zero (Cramer's determinants).

$D=\left|\begin{array}{c}12-3\\ 2\lambda 5\\ 143\mu \end{array}\right|$

Expanding along row 1:

$D=1(\lambda \mu -15)-2(2\mu -70)+(-3)(6-14\lambda )$

$=\lambda \mu -15-4\mu +140-18+42\lambda =\lambda \mu +42\lambda -4\mu +107$

Setting $D=0$: $\lambda \mu +42\lambda -4\mu +107=0\dots (\ast )$

Compute $D_2$ (replace column 2 with the RHS column $⟨2,5,33⟩$):

$D_2=\left|\begin{array}{c}12-3\\ 255\\ 1433\mu \end{array}\right|=1(5\mu -165)-2(2\mu -70)+(-3)(66-70)$

$=5\mu -165-4\mu +140+12=\mu -13$

$D_2=0\Rightarrow \mu =13$.

Compute $D_3$ similarly:

$D_3=\left|\begin{array}{c}122\\ 2\lambda 5\\ 14333\end{array}\right|=5\lambda +5$ (after expansion).

$D_3=0\Rightarrow \lambda =-1$.

Verify $(\ast )$: $(-1)(13)+42(-1)-4(13)+107=-13-42-52+107=0$. ✓

$\lambda +\mu =-1+13=12$.

Correct option: (4)