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

$\left|\begin{array}{ccc}k& 2& 1\\ 3& 0& -2\\ 3& -6k& -4\end{array}\right|=0$

$k=-2,k=\frac{1}{2}$

So far so good, but when subbing $-2$ for $k$

$-2x+2y+z=0\phantom{\rule{thickmathspace}{0ex}}...\phantom{\rule{thinmathspace}{0ex}}eqn\phantom{\rule{thinmathspace}{0ex}}1$

$3x+0y-2z=4\phantom{\rule{thickmathspace}{0ex}}...\phantom{\rule{thinmathspace}{0ex}}eqn\phantom{\rule{thinmathspace}{0ex}}2$

$3x+12y-4z=14\phantom{\rule{thickmathspace}{0ex}}...\phantom{\rule{thinmathspace}{0ex}}eqn\phantom{\rule{thinmathspace}{0ex}}3$

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 $3$

$\frac{5-2z}{2}+10-2z=14$

$\Rightarrow z=\frac{-1}{2},\phantom{\rule{thickmathspace}{0ex}}x=1\phantom{\rule{thickmathspace}{0ex}}and\phantom{\rule{thickmathspace}{0ex}}y=\frac{3}{4}$

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?