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{\bf Question}

Find the value of $k$ for which the equations
\begin{eqnarray*} x +2y & = & 0\\ 3x+ky-z & = & 0\\ 2x+5y-2z & = &
0 \end{eqnarray*}

have a solution, other than $x=y=z=0$. Find the solution set for
this value of $k$.

\medskip

{\bf Answer}

\begin{eqnarray*} x+2y & = & 0\ \ \ (1)\\ 3x+ky-z & = & 0\
\ \ (2)\\ 2x+5y-2z & = & 0\ \ \ (3) \end{eqnarray*}

Obviously $x=0\ y=0\ z=0$ are solutions.  Easiest way to do this
is to systematically eliminate; since $(1)$ is a "nice" equation:

\ \ \ $(1) \Rightarrow x=-2y$ $\downarrow$

$\Rightarrow (2)$ becomes $3(-2y)+ky-z=0$

\ \ \ $(3)$ becomes $2(-2y)+5y-2z=0$

Hence

\begin{eqnarray*} (k-6)y-z & = & 0 \ \ \ (4)\\ y-2z & = & 0 \ \ \
(5) \end{eqnarray*}

$(5) \Rightarrow y=2z$

Therefore $(k-6)(2z)-z=0$

\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $\Rightarrow (2k-13)z=0\ \ \ (6)$

Now from $(6)$ we could have $z=0 \Rightarrow y=0 \Rightarrow
x=0$. This isn't what we want. Another way to satisfy $(60$ is to
have $k=\ds\frac{13}{2}$. In this case $z$ could be anything, say
$z=\lambda$. Hence from $(5)$, $y=2\lambda$ and from $(1)$,
$x=-4\lambda$. Hence the solution is
$$x=-4\lambda,y=2\lambda,z=\lambda$$

or $-\ds\frac{x}{4}=\ds\frac{y}{2}=z$, the equation of a line in
3-D.

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