Solutions of a system of linear equations with a parameter I've come across a linear algebra problem that I'm not sure how to solve. It's a generic problem - I have to find the solutions of a system of equations dependent on a parameter. So, my first though was to check for which values of the parameter the matrix on the left is not invertible (1 and -2). I think that for these values there are no solutions. 
Is that enough, or should I check the rank of the augmented matrix and use the Rouché–Capelli theorem (for which values of the parameter the planes meet at a point, a line or a plane)? 
Thanks,
$$\begin{pmatrix}
\lambda & 1 & 1\\
1 & \lambda & 1\\
1 & 1 & \lambda
\end{pmatrix}
\begin{pmatrix}
  x \\  y \\  z
\end{pmatrix}=
\begin{pmatrix}
  1 \\  \lambda \\  \lambda^2
\end{pmatrix}
$$ 

$$\left(\begin{array}{ccc|c}
  \lambda & 1 & 1 & 1 \\
  1 & \lambda & 1 & \lambda \\
  1 & 1 & \lambda & \lambda^2
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  \lambda+2 & \lambda+2 & \lambda+2 & 1+\lambda+\lambda^2 \\
  1 & \lambda & 1 & \lambda \\
  1 & 1 & \lambda & \lambda^2
\end{array}\right)\overset{(1)}\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 1 & \frac{1+\lambda+\lambda^2}{\lambda+2} \\
  1 & \lambda & 1 & \lambda \\
  1 & 1 & \lambda & \lambda^2
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 1 & \frac{1+\lambda+\lambda^2}{\lambda+2} \\
  0 & \lambda-1 & 0 & \frac{\lambda-1}{\lambda+2} \\
  0 & 0 & \lambda-1 & \frac{\lambda^3+\lambda^2-\lambda-1}{\lambda+2}
\end{array}\right)\overset{(2)}\sim
\left(\begin{array}{ccc|c}
  1 & 1 & 1 & \frac{1+\lambda+\lambda^2}{\lambda+2} \\
  0 & 1 & 0 & \frac{1}{\lambda+2} \\
  0 & 0 & 1 & \frac{(\lambda+1)^2}{\lambda+2}
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  1 & 0 & 0 & \frac{-\lambda-1}{\lambda+2} \\
  0 & 1 & 0 & \frac{1}{\lambda+2} \\
  0 & 0 & 1 & \frac{(\lambda+1)^2}{\lambda+2}
\end{array}\right)\sim
$$
The step (1) is only valid for $\lambda\ne -2$.
The step (2) is only valid for $\lambda\ne 1$.
For values $\lambda\ne1,-2$ we get that there is only one solution $x_1=\frac{-\lambda-1}{\lambda+2}$, $x_2=\frac{1}{\lambda+2}$, $x_3=\frac{(\lambda+1)^2}{\lambda+2}$.
We can do the sanity check.
$\lambda x_1 + x_2 + x_3 = \frac{\lambda(-\lambda-1)+1+(\lambda+1)^2}{\lambda+2} = \frac{-\lambda^2-\lambda+1+\lambda^2+2\lambda+1}{\lambda+2} = \frac{\lambda+2}{\lambda+2} = 1$
$x_1 + \lambda x_2 + x_3 = \frac{(-\lambda-1)+\lambda+(\lambda+1)^2}{\lambda+2} = \frac{\lambda^2+2\lambda}{\lambda+2} = \lambda$
$x_1 + x_2 + \lambda x_3 = \frac{(-\lambda-1)+1+\lambda(\lambda+1)^2}{\lambda+2} = \frac{-\lambda+\lambda^3+2\lambda^2+\lambda}{\lambda+2} = \frac{\lambda^3+2\lambda^2}{\lambda+2}=\lambda^2$
Notice that the check works for $\lambda=1$. So we know that there is at least one solution for $\lambda=1$.
For $\lambda=1$
We get the system
$$\left(\begin{array}{ccc|c}
  1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1 \\
  1 & 1 & 1 & 1
\end{array}\right)$$
which has solutions $x_1=1-s-t$, $x_2=s$, $x_3=t$ (where $s$, $t$ are arbitrary).
For $\lambda=-2$
$\left(\begin{array}{ccc|c}
  -2 & 1 & 1 & 1 \\
  1 & -2 & 1 & -2 \\
  1 & 1 & -2 & 4
\end{array}\right)\sim
\left(\begin{array}{ccc|c}
  -2 & 1 & 1 & 1 \\
  1 & -2 & 1 & -2 \\
  0 & 0 & 0 & 3
\end{array}\right)$.
So in this case there is no solution. 
