If the system of linear equations, $x+y+z=6, x+2 y+3 z=10,3 x+2 y+\lambda z=\mu$ has more than two s

Question

If the system of linear equations, $x+y+z=6, x+2 y+3 z=10,3 x+2 y+\lambda z=\mu$ has more than two solutions, then $\mu-\lambda^{2}$ is equal to

Professor PC Thomas Classes · Answer

(13) : The given system of equations are $$\begin{equation*} x+y+z=6 	ag{i} \end{equation*}$$$$\begin{equation*} x+2 y+3 z=10 	ag{ii} \end{equation*}$$$$\begin{equation*} 3 x+2 y+\lambda z=\mu 	ag{iii} \end{equation*}$$ If $z=0$, then from (i) and (ii), we have\ $x+y=6$ and $x+2 y=10$\ $\Rightarrow \quad y=4, x=2$ i.e., solution is $(2,4,0)$\ If $y=0$, then from (i) and (ii), we have\ $x+z=6$ and $x+3 z=10$\ $\Rightarrow \quad z=2, x=4$ i.e., solution is $(4,0,2)$\ So, $\quad 3 x+2 y+\lambda z=\mu$ must pass through $(2,4,0)$\ and $(4,0,2)$.\ $\Rightarrow 6+8=\mu \Rightarrow \mu=14$ and $12+2 \lambda=\mu$\ $\Rightarrow \quad 12+2 \lambda=14 \Rightarrow \lambda=1$\ $	herefore \quad \mu-\lambda^{2}=14-1=13$\