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\begin{document}

{\bf Question}

Consider the following set of simultaneous equations

\begin{eqnarray*} x-y-z & = & 0\\ 3x+y+2z & = & 6\\ 2x+2y+kz & = &
2 \end{eqnarray*}

\begin{description}
\item[(a)]
If $h=1$, find the solution by matrix inversion.

\item[(b)]
If $k=3$, show that the equations do not have a unique solution.
(Do NOT attempt to find the solution set.)
\end{description}
\medskip

{\bf Answer}
\begin{description}
\item[(a)]
$k=1$:

\begin{eqnarray*} x-y-z & = & 0\\ 3x+y+2z & = & 6\\2x+2y+z & = & 2
\end{eqnarray*}

$\equiv \left(\begin{array}{ccc} 1 & -1 & -1\\ 3 & 1 & 2\\ 2 & 2 &
1 \end{array}
\right)\left(\begin{array}{c}x\\y\\z\end{array}\right)=\left(\begin{array}{c}0\\6\\2
\end{array} \right)$

\ \ \ \ \ \ \ \ \ \ \ ${\bf{A}} \ \ \ \ \ \ \cdot \ \ \ \ \ \ \
{\bf{X}}\ \ \ =\ \ \ {\bf{K}}$

Require ${\bf{A}}^{-1}$, since ${\bf{X}}={\bf{A}}^{-1}{\bf{K}}$

Step (iii)

\begin{eqnarray*} \bigtriangleup & = & det A = \left|\begin{array}{ccc} 1 & -1
& -1\\ 3 & 1 & 2\\ 2 & 2 & 1 \end{array} \right|\\ & = &
1\left|\begin{array}{cc}1 & 2\\ 2 & 1 \end{array}
\right|-(-1)\left|\begin{array}{cc}1 & 2\\ 2 & 1 \end{array}
\right| -1\left|\begin{array}{cc}1 & 2\\ 2 & 1
\end{array} \right|\\ & = & 1 \times(1-4)+1 \times(3-4)-1 \times
(6-2)\\ & = & -3-1-4\\ & = & -8 \end{eqnarray*}

Step (i):

Cofactors of the matrix are given by:

$\left(\begin{array} {ccc} A_{11} & A_{12} & A_{13}\\ A_{21} &
A_{22} & A_{23}\\ A_{31} & A_{32} & A_{33}
\end{array} \right)=\left(\begin{array} {ccc} 1 & -1 & -1\\ 3 & 1 & 2\\ 2 & 2 & 1
\end{array} \right)$

cofactor of $A_{11}=+\left|\begin{array} {cc} 1 & 2\\2 & 1
\end{array} \right|=-3$

cofactor of $A_{12}=-\left|\begin{array} {cc} 3 & 2\\2 & 1
\end{array} \right|=+1$

cofactor of $A_{13}=+\left|\begin{array} {cc} 3 & 1\\2 & 2
\end{array} \right|=+4$

cofactor of $A_{21}=-\left|\begin{array} {cc} -1 & -1\\2 & 1
\end{array} \right|=-1$

cofactor of $A_{22}=+\left|\begin{array} {cc} 1 & -1\\2 & 1
\end{array} \right|=+3$

cofactor of $A_{23}=-\left|\begin{array} {cc} 1 & -1\\2 & 2
\end{array} \right|=-4$

cofactor of $A_{31}=+\left|\begin{array} {cc} -1 & -1\\1 & 2
\end{array} \right|=-1$

cofactor of $A_{32}=-\left|\begin{array} {cc} 1 & -1\\3 & 2
\end{array} \right|=-5$

cofactor of $A_{33}=+\left|\begin{array} {cc} 1 & -1\\3 & 1
\end{array} \right|=+4$

Matrix of cofactors is thus:

$$\left(\begin{array} {ccc} -3 & 1 & 4\\ -1 & 3 & -4\\ -1 & -5 & 4
\end{array} \right)$$

Step (ii):

Transpose to get $adj A$

$$adj\ A=\left(\begin{array} {ccc} -3 & -1 & -1\\ 1 & 3 & -5\\ 4 &
-4 & 4 \end{array} \right)$$

Step (iii):

$det A=-8$

Step(iv):
\begin{eqnarray*} {\bf{A}} & = & \ds\frac{adj\ A}{\det\ A}\\ & = &
-\ds\frac{1}{8}\left(\begin{array}{ccc} -3 & -1 & -1\\1 & 3 & -5\\
4 & -4 & 4 \end{array} \right) \end{eqnarray*}

Hence

${\bf{X}}=-\ds\frac{1}{8}\left(\begin{array}{ccc} -3 & -1 & -1\\ 1
& 3 & -5\\ 4 & -4 & 4 \end{array}
\right)\left(\begin{array}{c}0\\6\\2 \end{array} \right)$

$\left(\begin{array}{c}x\\y\\z
\end{array}\right)=-\ds\frac{1}{8}\left(\begin{array}{c}
-8\\8\\-16 \end{array} \right)=\left(\begin{array}{c}1\\-1\\2
\end{array} \right)$

\item[(b)]
when $k=3$

$$det\left(\begin{array}{ccc} 1 & -1 & -1\\ 3 & 1 & 2\\ 2 & 2 &
k\end{array}\right)=0$$

where $k=3$

Thus no inverse. Hence no unique solution.
\end{description}
\end{document}