For what values of $k$ in this set of linear equations $Ax = b$ has no solutions, an infinite number of solutions and an unique solution? For what values of $k$ in this set of linear equations $Ax = b$ has no solutions, an infinite number of solutions and a unique solutions? 
I know I want to be using Gaussian Elimination here, I've augmented the matrix 
and I'm perfectly familiar with ERO's and back-solving for systems without unknown constants but this is new to me.
\begin{array}{ccc|c}  
 2 & 2 & 0 & 2\\  
 0 & k & 1 & 1\\
 1 & 2 & k & 2
\end{array}
Would I try to be putting this into Row-Echelon form? I have an inkling by playing with it that $k = -1$ for no solutions and $k = 1$ for an infinite number of solutions.  I can't do the Gaussian steps properly with a $k$ involved to produce some decent working though. 
Thank you in advance for any help, solutions or tips. :)
 A: Hint :
In order to have unique solutions, the determinant should be nonzero :
$$ \det(A) = 0 \Leftrightarrow \begin{array}{|ccc|c}  
 2 & 2 & 0 & \\  
 0 & k & 1  \\
 1 & 2 & k 
\end{array} =0 \Leftrightarrow 2(k^2-1) = 0 \Leftrightarrow k = +-1$$
Now, by plugging $k=1$ to our matrix and doing a Reduced Echelon Form Transformation :
$$\left(
\begin{array}{cccc}
 1 & 0 & -1 & 0 \\
 0 & 1 & 1 & 1 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)$$
and by plugging $k=-1$, executing again a Reduced Echelon Form Transformation :
$$\left(
\begin{array}{cccc}
 1 & 0 & 1 & 0 \\
 0 & 1 & -1 & 0 \\
 0 & 0 & 0 & 0 \\
\end{array}
\right)$$
Can you now derive conclusions for inconsistent, unique solution and infinite solutions?
A: As first comment already said you can use determinantats to find which answerz will dire t to a determinant equals to zero, which means a matrix which no aolution or with infinite solution. Using determinant you fimd that 1 and -1  are k values for determinant to be equals 0.  Now if you try to use those two values and reduce the matrix to find the solution then you will find that k=1 lead to infinite solutions, since it leads to a system of three variables and two rows. On the other hand with k=-1 you will find at some point of resuction that two rows show contradictory information, for example one auggesting -x1 + x2 = 1 and the other one suggesting that -x1+x2=-2. In some way its like there are three rows but just two variables, since you have a row that you dont need in order to have a system of two rows two variables. 
