Determine the values of $\lambda$ for the following results Q:Determine the values of $\lambda$ such that the following systems of linear equations have $(i)$ no solution ,$(ii)$More than one solution ,$(iii)$A unique solution. 
\begin{array}{c}
x+\lambda y+z=1 \\ 
x+y+\lambda z=1 \\ 
\lambda x+y+z=1
\end{array} My Approach:I used augmented matrix to determine the values of $\lambda$ $$
   \left( \begin{array}{rrr|r}
  1 & \lambda & 1 & 1  \\
   0 & 1 & \frac{1}{1+\lambda} & \frac{1}{1+\lambda}  \\
   0 & 0 & \frac{-2-\lambda}{1+\lambda} & -\frac{1}{1+\lambda}  \\ 
   \end{array}
\right)$$From here It's easy to determine the values of $\lambda =-2$ for case $(i)$.But i don't understand what for case $(ii),(iii)$.Any hints or solution will be appreciated.
Thanks in advance.
 A: As I said, your approach seems to have an error.  I don't know where it came from since you didn't show how you got the augmented matrix.  I am presenting a different solution.
Summing all the equations, we get $(\lambda+2)(x+y+z)=3$.  Therefore, $\lambda=-2$ gives $0=3$, which is absurd.  So, $\lambda=-2$ lands you in case ($i$).  
For $\lambda=1$, you have three identical equations, which are all $x+y+z=1$.  Hence, $\lambda=1$ lands you in case ($ii$).
I claim that if $\lambda\neq 1,-2$, then you have case ($iii$).  We subtract the first equation from the second equation to get $(\lambda-1)(y-z)=0$.  Because $\lambda\neq 1$, we have $y=z$.  Similarly, $x=y$ (by subtracting the first and the last equation).  Therefore, $x=y=z=\frac{1}{\lambda+2}$, which is a valid expression since $\lambda\neq -2$.
A: A=$\begin{pmatrix}1 &\lambda & 1 & 1\\0 & 1-\lambda & \lambda-1 & 0\\0 & 0& 2-\lambda-\lambda^2 & 1-\lambda\end{pmatrix}$ is the correct matrix. You can't divide with 1+$\lambda$ because it isn't given anywhere that $\lambda\neq-1$.
Now for no solution $2-\lambda-\lambda^2=0$ and $1-\lambda\neq0$. So $\lambda =-2$  is correct.
For more than one solution rank(A)<3. So when $1-\lambda=0$ and $2-\lambda-\lambda^2=0$ more than one solution occurs. $\lambda=1$.
For unique solution, the matrix A must be invertible. So $det(A)\neq0$.
$(2-\lambda-\lambda^2)(1-\lambda)\neq0$
So $\lambda\neq1,-2$
