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{\bf Question}

For the following system of equations $$ \left(\begin{array}{ccc}
1&amp; 2 &amp; 1 \\ 1 &amp; 1 &amp; -1 \\ 0 &amp; -1 &amp; 2 \end{array}
\right)\left(\begin{array}{c} x\\y\\z\end{array} \right) = \left(
\begin{array}{c} 6 \\ 7 \\ 3 \end{array} \right)$$
\begin{description}
\item[(a)] Write down the matrix and the augmented matrix
\item[(b)] Find the rank of both by the elimination method
\item[(c)] Use this information to determine whether the equations
have a solution, and if they do how many free variables there are.
\item[(d)] If they do have a solution, find it, and confirm that
indeed it has the right number of free variables.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
$A = \left(\begin{array}{ccc} 1&amp; 2 &amp; 1 \\ 2 &amp; 1 &amp; -1 \\ 1 &amp; -1 &amp; 0
\end{array} \right) \hspace{.3in} A:b = \left(\begin{array}{cccc}
 1&amp; 2 &amp; 1 &amp; 6\\ 2 &amp; 1 &amp; -1 &amp; 7 \\ 1 &amp; -1 &amp; 0&amp;3 \end{array} \right)$

\item[(b)] Use elimination method to find rank

$\left(\begin{array}{cccc} 1&amp; 2 &amp; 1 &amp; 6\\ 2 &amp; 1 &amp; -1 &amp; 7 \\ 1 &amp; -1
&amp; 0&amp;3 \end{array} \right) \rightarrow\ ({\rm exchange\ rows\ 1,
2})
$

$\left(\begin{array}{cccc}  2 &amp; 1 &amp; -1 &amp; 7 \\ 1&amp; 2 &amp; 1 &amp; 6\\ 1 &amp;
-1 &amp; 0&amp;3 \end{array} \right) \begin{array}{c}\rightarrow\ ({\rm
row}\  2 \rightarrow 2 {\rm row}\  2 - {\rm row}\  1) \\
\hspace{.3in} ({\rm row}\  3 \rightarrow  2 {\rm row}\  3 - {\rm
row}\ 1)\end{array}$

$\left(\begin{array}{cccc} 2&amp; 1 &amp; -1 &amp; 7\\ 0 &amp; 3 &amp; 3 &amp; 5 \\ 0 &amp; -3
&amp; 1&amp;-1 \end{array} \right) \rightarrow\ ({\rm row}\  3 \rightarrow
{\rm row}\  3 + {\rm row}\  2)$

$\left(\begin{array}{cccc} 2&amp; 1 &amp; -1 &amp; 7\\ 0 &amp; 3 &amp; 3 &amp; 5 \\ 0 &amp; 0
&amp; 4&amp;4 \end{array} \right)$

Hence both $r(A) = r(A:b) = 3$

\item[(c)]
Hence equations do have a solution and since $r(A) = r(A:b)$, no.
of free parameters = no of unknowns $-\ r(A) = 3-3 = 0$

\item[(d)]
Equations are \begin{eqnarray*} 2x + y - z &amp; = &amp; 7 \\ 3y + 3z &amp; =
&amp; 5 \\ 4z &amp; = &amp; 4\end{eqnarray*}

Let $z = 1 \Rightarrow y = \frac{2}{3} \Rightarrow 2x = 7 +z -y =
7+1-\frac{2}{3}=\frac{22}{3} \Rightarrow x = \frac{11}{3}$ and
${\bf x} = \left(\begin{array}{c} \frac{11}{3}
\\ \frac{2}{3}
\\ 1
\end{array} \right)$ with no free variable as expected.
\end{description}



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