Simultaneous Equations That Should Be Inconsistent Has a Unique Solution Find the values of $k$ for which the simultaneous equations do not have a unique solution for $x, y$ and $z$.
Also show that when $k = -2$ the equations are inconsistent
$$kx + 2y +z =0$$
$$3x + 0y -2z =4$$
$$3x - 6ky -4z =14$$
Using a determinant and setting to zero, then solving the quadratic gives
$$
\begin{vmatrix}
k & 2 & 1\\
3 & 0 & -2\\
3 & -6k & -4\\
\end{vmatrix}
=0
$$
$$
k=-2, k=\frac{1}{2}
$$
So far so good, but when subbing $-2$ for $k$
$$-2x + 2y +z =0 \;...\,eqn \,1$$
$$3x + 0y -2z =4 \;...\,eqn \,2$$
$$3x +12y -4z =14 \;...\,eqn \,3$$
subbing eqn $2$ from eqn $3$ gives 
$$
12y-2z=10$$
$$\Rightarrow
y = \frac{5 +z}{6}
$$
Subbing for $y$ into eqn $1$
$$
-2x+\frac{5+z}{3}-z=0
$$
$$
\Rightarrow x = \frac{5-2z}{6}
$$
Subbing for $y$ and $x$ into eqn $3$
$$\frac{5-2z}{2}+10 -2z =14 $$
$$\Rightarrow z = \frac{-1}{2},\;
x = 1 \; and \; y = \frac{3}{4}$$
These values for $x,y$ and $z$ seem to prove unique solutions for these equations, yet from the determinant and also the question in the text book should they not be inconsistent?
Thanks in advance to anyone who helps with regards to this exercise from a slightly confused self learner!
 A: Firstly, a non-zero determinant doesn't imply there are no solutions.  The system of equations
$$
\left[
\begin{array}{ccc|c}
1 & 1 & 1 & 1 \\
2 & 2 & 2 & 2 \\
3 & 3 & 3 & 3 \\
\end{array}
\right]
$$
has an infinite number of solutions, for example, and the relevant determinant will be zero.  So, we can't be sure the system will be inconsistent from a zero determinant alone.
The matrix
$$
\left[
\begin{array}{ccc|c}
-2 & 2 & 1 & 0 \\
3 & 0 & -2 & 4 \\
3 & 12 & -4 & 14 \\
\end{array}
\right]
$$
has row echelon form
\begin{align*}
& \sim_{R_3 \gets R_3-R_2}
\left[
\begin{array}{ccc|c}
-2 & 2 & 1 & 0 \\
3 & 0 & -2 & 4 \\
0 & 12 & -2 & 10 \\
\end{array}
\right] \\
& \sim_{R_2 \gets R_2+\frac{3}{2}R_1}
\left[
\begin{array}{ccc|c}
-2 & 2 & 1 & 0 \\
0 & 3 & -0.5 & 4 \\
0 & 12 & -2 & 10 \\
\end{array}
\right] \\
& \sim_{R_3 \gets R_3-4R_2}
\left[
\begin{array}{ccc|c}
-2 & 2 & 1 & 0 \\
0 & 3 & -0.5 & 4 \\
0 & 0 & 0 & -6 \\
\end{array}
\right] \\
\end{align*}
which is inconsistent.  So, no solutions exist.
