Find the real values

Solution:

The given system of equations can be written as

$2 \lambda x-2 y+3 z=0$

$x+\lambda y+2 z=0$

$2 x+0 y+\lambda z=0$

The given system of equations will have non-trivial solution $s$ if $D=0$.

$\Rightarrow\left|\begin{array}{ccc}2 \lambda & -2 & 3 \\ 1 & \lambda & 2 \\ 2 & 0 & \lambda\end{array}\right|=0$

$\Rightarrow 2 \lambda\left(\lambda^{2}\right)+2(\lambda-4)+3(-2 \lambda)=0$

$\Rightarrow 2 \lambda^{3}-4 \lambda-8=0$

$\Rightarrow \lambda=2$

So, the given system of equations will have non-trivial solutions if $\lambda=2$.

Now, we shall find solutions for $\lambda=2$.

Replacing $z$ by $k$ in the first two equations, we get

$2 \lambda x-2 y=-3 k$

$x+\lambda y=-2 k$

Solving these by Cramer's rule, we get

$x=\frac{\left|\begin{array}{cc}-3 k & -2 \\ -2 k & \lambda\end{array}\right|}{\left|\begin{array}{cc}2 \lambda & -2 \\ 1 & \lambda\end{array}\right|}=\frac{-3 k \lambda-4 k}{2 \lambda^{2}+2}=\frac{-3 k(2)-4 k}{2(2)^{2}+2}=\frac{-6 k-4 k}{10}=-k$

$y=\frac{\left|\begin{array}{cc}2 \lambda & -3 k \\ 1 & -2 k\end{array}\right|}{\left|\begin{array}{cc}2 \lambda & -2 \\ 1 & \lambda\end{array}\right|}=\frac{-4 k \lambda+3 k}{2 \lambda^{2}+2}=\frac{-4 k(2)+3 k}{2(2)^{2}+2}=\frac{-5 k}{10}=\frac{-k}{2}$

Substituting these values of $x$ and $y$ in the third equation, we get

LHS $=2(-k)+0\left(-\frac{k}{2}\right)+2(k)=0=$ RHS

Thus,

$\lambda=2, x=-k, y=-\frac{k}{2}$ and $z=k \quad[k \in R]$