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$$x+3y+kz=a$$ $$2x+(2k+2)y+(3k-2)z=b$$ $$kx+(k+4)y+4z=c$$

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?

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    $\begingroup$ The second equation has $(3x-2)z$ which makes the system nonlinear. Did you mean $(3k-2)z$? $\endgroup$ – Nitish Feb 19 '14 at 21:57
  • $\begingroup$ @Nitish yes sorry ill correct it now! $\endgroup$ – Lauren Feb 19 '14 at 22:07
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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.

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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.

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