<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\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 &amp; = &amp; 4\\ x-2y-4z &amp; = &amp; 2\\2x+5y+3z &amp; =
&amp; -1 \end{eqnarray*}

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

\end{description}
\medskip

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

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

Thus in short,

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

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

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

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

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

and

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

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

Similarly from notes we have:

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

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

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

Similarly we have

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

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

${}$

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

Thus

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

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

and

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

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

\newpage
Now also

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

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

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

and finally,

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

\begin{eqnarray*} \left|\begin{array}{ccc} d_1 &amp; a_1 &amp; b_1\\ d_2 &amp; a_2 &amp;
b_2\\ d_3 &amp; a_3 &amp; b_3 \end{array} \right| &amp; = &amp;
\left|\begin{array}{ccc} 3 &amp; 4 &amp; 4\\ 1 &amp; 2 &amp; -2\\ 2 &amp; -1 &amp; 5
\end{array} \right|\\ &amp; = &amp; 3\left|\begin{array}{cc} 2 &amp; -2\\ -1 &amp; 5
\end{array} \right|-4\left|\begin{array}{cc} 1 &amp; -2\\ 2 &amp; 5
\end{array} \right|+4\left|\begin{array}{cc} 1 &amp; 2\\ 2 &amp; -1
\end{array} \right|\\ &amp; = &amp; 3(10-2)-4(5+4)+4(-1-4)\\ &amp; = &amp;
24-36-20\\ &amp; = &amp; \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}
</pre></body></html>