Find the real values of k in a linear systen that does not have a unique solution $$x+3y+kz=a$$
$$2x+(2k+2)y+(3k-2)z=b$$
$$kx+(k+4)y+4z=c$$
I have put them into a matrix and now am unsure where to go from here, do i make a RREF?
also how would i do it to make them consistent?
 A: So the matrix should just have the following form
$$Ax = b$$
where A is the matrix defining the system.  All of the x coefficients go in the first column y in the 2nd z in the third.  
$$ \left( \begin{array}{ccc}
1& 3 & k \\
2 & (2k+2) & (3k - 2) \\
k & (k+4) & 4 \end{array} \right) * \left( \begin{array}{c}
x \\
y  \\
z \end{array} \right) =  \left( \begin{array}{c}
a\\
b  \\
c \end{array} \right)$$
Take the Determinant and set it to zero.  That will give you k values that make the matrix linearly dependent which are not unique.
Wolfram Alpha
$$Det[A] = -12 + 7 k^2 - 2 k^3 = 0$$
solving for the above equation gives the following values for k:
Wolfram Alpha
$$k = 2$$
$${k }= {1\over4}(3 \pm   \sqrt{57})$$
Those are the values for k that define a system that has no unique solution.  All other k values do have a unique solution.   
A: You are looking for $k\in \mathbb{R}$ such that the three given lines (interpreted as vectors) are linear dependent. If they are linear dependent for some value $k$, then the system has no unique solution.
