JEE MAIN - Mathematics (2024 - 5th April Morning Shift - No. 10)

If the system of equations

$$\begin{array}{r} 11 x+y+\lambda z=-5 \\ 2 x+3 y+5 z=3 \\ 8 x-19 y-39 z=\mu \end{array}$$

has infinitely many solutions, then $$\lambda^4-\mu$$ is equal to :

51
45
47
49

Explanation

$$\begin{aligned} & 11 x+y+\lambda z=-5 \\ & 2 x+3 y+5 z=3 \\ & 8 x-19 y-39 z=\mu \\ & \Delta=0 \Rightarrow\left|\begin{array}{ccc} 11 & 1 & \lambda \\ 2 & 3 & 5 \\ 8 & -19 & -39 \end{array}\right|=0 \\ & 11(-39.3+19.5)-1(-39.2-40)+\lambda(-38-24)=0 \\ & =11(-117+95)-1(-118)-62 \lambda=0 \\ & =-242+118=62 \lambda \\ & \Rightarrow \lambda=-2 \\ & \Delta 2=0 \\ & \Rightarrow\left|\begin{array}{ccc} 11 & 1 & -5 \\ 2 & 3 & 3 \\ 8 & -19 & \mu \end{array}\right|=0 \end{aligned}$$

$$\begin{aligned} & 11(3 \mu+57)-1(2 \mu-24)-5(-38-24)=0 \\ & 33 \mu+627-2 \mu+24+310=0 \\ & \mu=-31 \\ & \Rightarrow \lambda^4-31 \\ & =16+31 \\ & =47 \end{aligned}$$

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