\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\parindent=0pt
\begin{document}

{\bf Question}

Solve the following simultaneous equations and identify the
intersection points of the corresponding planes

\begin{description}
\item[(i)]
\begin{eqnarray*} 3x+4y+2z & = & 4\\ x-2y-4z & = & 2\\2x+5y+3z & =
& -1 \end{eqnarray*}

\item[(ii)]
\begin{eqnarray*} 4x+4y+2z & = & 3\\ 2x-2y-4z & = & 1\\-x+5y+3z & =
& 2 \end{eqnarray*}

\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(i)]
From lecture notes:

$\left.\begin{array} {rcl} 3x+4y+2z & = & 4\\ x -2y-4z & = & 2\\
2x+5y+3z & = & -1 \end{array} \right\} cf. \left\{\begin{array}
{rcl} a_1x+b_1y+c_1z=d_1\\ a_2x+b_2y+c_2z=d_2\\ a_3x+b_3y+c_3z=d_3
\end{array} \right.$

Thus in short,

$\left(\begin{array}{cccc} a_1 & b_1 & c_1 & d_1\\ a_2 & b_2 & c_2
& d_2\\ a_3 & b_3 & c_3 & d_3 \end{array} \right) =
\left(\begin{array}{cccc} 3 & 4 & 2 & 4\\ 1 & -2 & -4 & 2\\ 2 & 5&
3 & -1 \end{array} \right)$

i.e., $a_1=3,\ b_2=-2$ etc.

Now equations ($4.15 a,b,c$) gives the $x$ solution as:

$x=\ds\frac{\left|\begin{array}{ccc} d_1 & b_1 & c_1\\ d_2 & b_2 &
c_2\\ d_3 & b_3 & c_3
\end{array} \right|}{\left|\begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 &
c_3 \end{array} \right|}$

\newpage
Calculate
\begin{eqnarray*} \left|\begin{array}{ccc} d_1 & b_1 & c_1\\ d_2 & b_2 &
c_2\\ d_3 & b_3 & c_3 \end{array} \right| & = &
\left|\begin{array}{ccc} 4 & 4 & 2\\ 2 & -2 & -4\\ -1 & 5 & 3
\end{array} \right|\\ & = & 4\left|\begin{array}{cc} -2 & -4\\ 5 & 3
\end{array} \right|-4\left|\begin{array}{cc} 2 & -4\\ -1 & 3
\end{array} \right|+2\left|\begin{array}{cc} 2 & -2\\ -1 & 5
\end{array} \right|\\ & = & 4(-6+20)-4(6-4)+2(10-2)\\ & = &
56-8+16\\ & = & \un{64} \end{eqnarray*}

and

\begin{eqnarray*} \left|\begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 &
c_2\\ a_3 & b_3 & c_3 \end{array} \right| & = &
\left|\begin{array}{ccc} 3 & 4 & 2\\ 1 & -2 & -4\\ 2 & 5 & 3
\end{array} \right|\\ & = & 3\left|\begin{array}{cc} -2 & -4\\ 5 & 3
\end{array} \right|-4\left|\begin{array}{cc} 1 & -4\\ 2 & 3
\end{array} \right|+2\left|\begin{array}{cc} 1 & -2\\ 2 & 5
\end{array} \right|\\ & = & 3(-6+20)-4(3+8)+2(5+4)\\ & = &
42-44+18\\ & = & \un{16} \end{eqnarray*}

Thus $x=\ds\frac{64}{16}=\un{4}$

Similarly from notes we have:

$y=\ds\frac{\left|\begin{array}{ccc} d_1 & c_1 & a_1\\ d_2 & c_2 &
a_2\\ d_3 & c_3 & a_3
\end{array} \right|}{\left|\begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 &
c_3 \end{array} \right|}=16$ from above

\newpage
\begin{eqnarray*} \left|\begin{array}{ccc} d_1 & c_1 & a_1\\ d_2 & c_2 &
a_2\\ d_3 & c_3 & a_3 \end{array} \right| & = &
\left|\begin{array}{ccc} 4 & 2 & 3\\ 2 & -4 & 1\\ -1 & 3 & 2
\end{array} \right|\\ & = & 4\left|\begin{array}{cc} -4 & 1\\ 3 & 2
\end{array} \right|-2\left|\begin{array}{cc} 2 & 1\\ -1 & 2
\end{array} \right|+3\left|\begin{array}{cc} 2 & -4\\ -1 & 3
\end{array} \right|\\ & = & 4(-8-3)-2(4+1)+3(6-4)\\ & = &
-44-10+6\\ & = & \un{-48} \end{eqnarray*}

Thus $y=\ds\frac{-48}{16}=\un{-3}$

Similarly we have

$z=\ds\frac{\left|\begin{array}{ccc} d_1 & a_1 & b_1\\ d_2 & a_2 &
b_2\\ d_3 & a_3 & b_3
\end{array} \right|}{\left|\begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 &
c_3 \end{array} \right|}=16$ from above

\begin{eqnarray*} \left|\begin{array}{ccc} d_1 & a_1 & b_1\\ d_2 & a_2 &
b_2\\ d_3 & a_3 & b_3 \end{array} \right| & = &
\left|\begin{array}{ccc} 4 & 3 & 4\\ 2 & 1 & -2\\ -1 & 2 & 5
\end{array} \right|\\ & = & 4\left|\begin{array}{cc} 1 & -2\\ 2 & 5
\end{array} \right|-3\left|\begin{array}{cc} 2 & -2\\ -1 & 5
\end{array} \right|+4\left|\begin{array}{cc} 2 & 1\\ -1 & 2
\end{array} \right|\\ & = & 4(5+4)-3(10-2)+4(4+1)\\ & = &
36-24+20\\ & = & \un{32} \end{eqnarray*}

Thus $z=\ds\frac{32}{16}=\un{2}$

Thus $x=4,\ y=-3,\ z=2$

are the solutions of this set of simultaneous equations. Hence the
3-planes represented by these 3 equations intersect at a
\un{point}, P.

PICTURE \vspace{1in}


\item[(ii)]

${}$

$\left(\begin{array}{cccc} a_1 & b_1 & c_1 & d_1\\ a_2 & b_2 & c_2
& d_2\\ a_3 & b_3 & c_3 & d_3 \end{array} \right) =
\left(\begin{array}{cccc} 4 & 4 & 2 & 3\\ 3 & -2 & -4 & 1\\ -1 & 5
& 3 & 2 \end{array} \right)$

Thus

$x=\ds\frac{\left|\begin{array}{ccc} d_1 & b_1 & c_1\\ d_2 & b_2 &
c_2\\ d_3 & b_3 & c_3
\end{array} \right|}{\left|\begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 &
c_3 \end{array} \right|}$

Calculate
\begin{eqnarray*} \left|\begin{array}{ccc} d_1 & b_1 & c_1\\ d_2 & b_2 &
c_2\\ d_3 & b_3 & c_3 \end{array} \right| & = &
\left|\begin{array}{ccc} 3 & 4 & 2\\ 1 & -2 & -4\\ 2 & 5 & 3
\end{array} \right|\\ & = & 3\left|\begin{array}{cc} -2 & -4\\ 5 & 3
\end{array} \right|-4\left|\begin{array}{cc} 1 & -4\\ 2 & 3
\end{array} \right|+2\left|\begin{array}{cc} 1 & -2\\ 2 & 5
\end{array} \right|\\ & = & 3(-6+20)-4(3+8)+2(5+4)\\ & = &
42-44+18\\ & = & \un{16} \end{eqnarray*}

and

\begin{eqnarray*} \left|\begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 &
c_2\\ a_3 & b_3 & c_3 \end{array} \right| & = &
\left|\begin{array}{ccc} 4 & 4 & 2\\ 2 & -2 & -4\\ -1 & 5 & 3
\end{array} \right|\\ & = & 4\left|\begin{array}{cc} -2 & -4\\ 5 & 3
\end{array} \right|-4\left|\begin{array}{cc} 2 & -4\\ -1 & 3
\end{array} \right|+2\left|\begin{array}{cc} 2 & -2\\ -1 & 5
\end{array} \right|\\ & = & 4(-6+20)-4(6-4)+2(10-2)\\ & = &
56-8+16\\ & = & \un{64} \end{eqnarray*}

Thus $x=\ds\frac{16}{64}=\un{\ds\frac{1}{4}}$

\newpage
Now also

$y=\ds\frac{\left|\begin{array}{ccc} d_1 & c_1 & a_1\\ d_2 & c_2 &
a_2\\ d_3 & c_3 & a_3
\end{array} \right|}{\left|\begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 &
c_3 \end{array} \right|}=64$ from above

\begin{eqnarray*} \left|\begin{array}{ccc} d_1 & c_1 & a_1\\ d_2 & c_2 &
a_2\\ d_3 & c_3 & a_3 \end{array} \right| & = &
\left|\begin{array}{ccc} 3 & 2 & 4\\ 1 & -4 & 2\\ 2 & 3 & -1
\end{array} \right|\\ & = & 3\left|\begin{array}{cc} -4 & 2\\ 3 & -1
\end{array} \right|-2\left|\begin{array}{cc} 1 & 2\\ 2 & -1
\end{array} \right|+4\left|\begin{array}{cc} 1 & -4\\ 2 & 3
\end{array} \right|\\ & = & 3(+4-6)-2(-1-4)+4(3+8)\\ & = &
-6+10+44\\ & = & \un{48} \end{eqnarray*}

Thus $y=\ds\frac{48}{64}=\un{\ds\frac{3}{4}}$

and finally,

$z=\ds\frac{\left|\begin{array}{ccc} d_1 & a_1 & b_1\\ d_2 & a_2 &
b_2\\ d_3 & a_3 & b_3
\end{array} \right|}{\left|\begin{array}{ccc} a_1 & b_1 & c_1\\ a_2 & b_2 & c_2\\ a_3 & b_3 &
c_3 \end{array} \right|}=64$ from above

\begin{eqnarray*} \left|\begin{array}{ccc} d_1 & a_1 & b_1\\ d_2 & a_2 &
b_2\\ d_3 & a_3 & b_3 \end{array} \right| & = &
\left|\begin{array}{ccc} 3 & 4 & 4\\ 1 & 2 & -2\\ 2 & -1 & 5
\end{array} \right|\\ & = & 3\left|\begin{array}{cc} 2 & -2\\ -1 & 5
\end{array} \right|-4\left|\begin{array}{cc} 1 & -2\\ 2 & 5
\end{array} \right|+4\left|\begin{array}{cc} 1 & 2\\ 2 & -1
\end{array} \right|\\ & = & 3(10-2)-4(5+4)+4(-1-4)\\ & = &
24-36-20\\ & = & \un{-32} \end{eqnarray*}

Hence $z=\ds\frac{-32}{64}=\un{-\ds\frac{1}{2}}$

Thus $x=\ds\frac{1}{4},\ y=\ds\frac{3}{4},\ z=\ds\frac{-1}{2}$

are the solutions of this set of simultaneous equations. Hence the
3-planes represented by these 3 equations intersect at a point.
\end{description}
\end{document}
