Question

Find the real values of $\lambda$ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions

$$\begin{gathered} 2\lambda x-2y+3z=0 \\ x+\lambda y+2z=0 \\ 2x+\lambda z=0 \end{gathered}$$

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as} \\ 2\mathrm{\lambda }x-2y+3z=0 \\ x+\lambda y+2z=0 \\ 2x+0y+\lambda z=0 \\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{will}\mathrm{have}\mathrm{non}-\mathrm{trivial}\mathrm{solution}s\mathrm{if}D=0. \\ \Rightarrow \left|\begin{array}{ccc}2\lambda & -2& 3\\ 1& \lambda & 2\\ 2& 0& \lambda \end{array}\right|=0 \\ \Rightarrow 2\lambda \left({\lambda }^2\right)+2(\lambda -4)+3(-2\lambda )=0 \\ \Rightarrow 2{\lambda }^3-4\lambda -8=0 \\ \Rightarrow \lambda =2 \\ \mathrm{So},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{will}\mathrm{have}\mathrm{non}-\mathrm{trivial}\mathrm{solutions}\mathrm{if}\lambda =2. \\ \mathrm{Now},\mathrm{we}\mathrm{shall}\mathrm{find}\mathrm{solutions}\mathrm{for}\lambda =2. \\ \mathrm{Replacing}z\mathrm{by}k\mathrm{in}\mathrm{the}\mathrm{first}\mathrm{two}\mathrm{equations},\mathrm{we}\mathrm{get} \\ 2\lambda x-2y=-3k \\ x+\lambda y=-2k \\ \mathrm{Solving}\mathrm{these}\mathrm{by}\mathrm{Cramer}\text{'}\mathrm{s}\mathrm{rule},\mathrm{we}\mathrm{get} \\ x=\frac{\left|\begin{array}{cc}-3k& -2\\ -2k& \lambda \end{array}\right|}{\left|\begin{array}{cc}2\lambda & -2\\ 1& \lambda \end{array}\right|}=\frac{-3k\lambda -4k}{2{\lambda }^2+2}=\frac{-3k\left(2\right)-4k}{2(2{)}^2+2}=\frac{-6k-4k}{10}=-k \\ y=\frac{\left|\begin{array}{cc}2\lambda & -3k\\ 1& -2k\end{array}\right|}{\left|\begin{array}{cc}2\lambda & -2\\ 1& \lambda \end{array}\right|}=\frac{-4k\lambda +3k}{2{\lambda }^2+2}=\frac{-4k\left(2\right)+3k}{2(2{)}^2+2}=\frac{-5k}{10}=\frac{-k}{2} \\ \mathrm{Substituting}\mathrm{these}\mathrm{values}\mathrm{of}x\mathrm{and}y\mathrm{in}\mathrm{the}\mathrm{third}\mathrm{equation},\mathrm{we}\mathrm{get} \\ \mathrm{LHS}=2(-k)+0(-\frac{k}{2})+2\left(k\right)=0=\mathrm{RHS} \\ \mathrm{Thus}, \\ \mathrm{\lambda }=2,x=-k,y=-\frac{k}{2}\mathrm{and}z=k\left[k\in R\right] \end{gathered}$$