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{\bf Question}

By evaluating the relevant determinant, investigate for various
values of $a$ and $b$ the type of solution set of the equations

\begin{eqnarray*} x+3y-2z & = & 7\\ax+6y-4z & = & 2-3b\\ 2x+6y+bz
& = & 14 \end{eqnarray*}

\medskip

{\bf Answer}

The equations can be written as:

$$\left(\begin{array}{ccc} 1 & 3 & -2\\ a & 6 & -4\\ 2 & 6 & b
\end{array} \right)\left(\begin{array}{c}x\\y\\z
\end{array}\right)=\left(\begin{array}{c} 7 \\ 2-3b \\ 14 \end{array}
\right)$$

whether we solve by inverting the coefficient matrix or with a
determinant method (Cramer's rule), we have to work out its
determinant.

If this determinant=0 we don't have an inverse, or alternatively,

Cramer's rule will fail.

Thus we need to work out

\begin{eqnarray*} \left|\begin{array}{ccc} 1 & 3 & -2\\ a & 6 & -4\\ 2 & 6 & b
\end{array} \right| & = & 1 \times \left|\begin{array}{cc} 6 & -4
\\ 6 & b\end{array}\right|-3\left|\begin{array}{cc} a & -4
\\ 2 & b\end{array}\right|+(-2)\left|\begin{array}{cc} a & 6
\\ 2 & 6\end{array}\right|\\ & = & 6b+24-3ab-24-12a+24\\ & = &
24-12a+6b-3ab\\ & = & 3(8-4a+2b-ab)\\ & = & 3(2-a)(4+b)
\end{eqnarray*}

So \begin{description}
\item[(i)]
If $a \ne 2$ and $b \ne -4$ there is a unique solution, since $det
\ne 0$. The planes meet at a point

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\item[(ii)]
If $a = 2$ and $b \ne -4$ then $det =0$. The system is

\begin{eqnarray*} x+3y-2z & = & 7\ \ \ (1)\\ 2x+6y-4z & = & 2-3b\
\ \ (2)\\ 2x+6y+bz & = & 14\ \ \ (3) \end{eqnarray*}

$(2) \Rightarrow x+3y-2z = 1-3\ds\frac{b}{2}$

cf $(1) \Rightarrow x+3y-2z=7$

These are two parallel planes if $1-3\ds\frac{b}{2} \ne 7
\Rightarrow b \ne 4$

So they're parallel and non intersecting:

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\end{picture}

\item[(iii)]
If $b=-4$ but $a \ne 2$ we have

\begin{eqnarray*} x+3y-2z & = & 7\\ ax+6y-4z & = & 14\\ 2x+6y-4z &
= & 14 \end{eqnarray*}

$(3)$ is twice $(1)$

$(2)$ and $(3)$ are not the same since $a \ne 2$

We have two coincident planes with an intersecting plane.

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\put(3.1,4){\makebox(0,0)[l]{intersecting plane}}
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Thus the solution is a \un{line} of points not a single one.

\item[(iv)]
If $a=2$ and $b=-4$ we get

$\left.\begin{array} {rcl} x+3y-2z & = & 7\\ ax+6y-4z & = & 14\\
2x+6y-4z & = & 14 \end{array} \right\}$ all the same equation!!!

We have 3-coincident planes.

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\put(4.5,2.2){\makebox(0,0)[l]{3 coincident planes}}
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So a \un{plane} of $x,y,z$ solutions, not a single point.
\end{description}

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