JEE MAIN - Mathematics (2024 - 9th April Morning Shift - No. 17)
Let $$\lambda, \mu \in \mathbf{R}$$. If the system of equations
$$\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned}$$
has infinitely many solutions, then $$\mu+2 \lambda$$ is equal to :
Explanation
$$\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned}$$
$$\begin{aligned} & {\left[\begin{array}{ccc} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 3 \\ 2 \\ \mu \end{array}\right]} \\ & A X=B \\ & X=A^{-1} B \\ & X=\frac{\operatorname{adj} A}{|A|} B \end{aligned}$$
For Infinitely many solution
$$\begin{aligned} & |A|=0 \text { and }(\operatorname{adj} A) B=0 \\ & \left|\begin{array}{ccc} 3 & 5 & \lambda \\ 7 & 11 & -9 \\ 97 & 155 & -189 \end{array}\right|=0 \\ & -2052+2250+18 \lambda=0 \\ & \Rightarrow \lambda=-11 \\ & \operatorname{adj} A=\left[\begin{array}{ccc} -684 & -760 & 76 \\ 450 & 500 & -50 \\ 18 & 20 & -2 \end{array}\right] \\ & (\operatorname{adj} A) B=\left[\begin{array}{ccc} 684 & -760 & 76 \\ 450 & 500 & -50 \\ 18 & 20 & -2 \end{array}\right]\left[\begin{array}{l} 3 \\ 2 \\ \mu \end{array}\right]=\left[\begin{array}{l} 0 \\ 0 \\ 0 \end{array}\right] \\ & \Rightarrow 54+40-2 \mu=0 \\ & \Rightarrow 2 \mu=94 \\ & \Rightarrow \mu=47 \\ & \Rightarrow \mu+2 \lambda=25 \end{aligned}$$
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