Question

Find the real values of λ for which the following system of linear equations has non-trivial solutions. Also, find the non-trivial solutions
\[2 \lambda x - 2y + 3z = 0\] 
\[ x + \lambda y + 2z = 0\] 
\[ 2x + \lambda z = 0\]

 

Answer 1

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