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{\bf Question}

For the following system of equations $$ \left(\begin{array}{ccc}
1& 2 & 1 \\ 1 & 1 & -1 \\ 0 & 1 & 2 \end{array}
\right)\left(\begin{array}{c} x\\y\\z\end{array} \right) = \left(
\begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right)$$
\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}


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{\bf Answer}

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

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

$\left(\begin{array}{cccc} 1& 2 & 1 & 0\\ 1 & 1 & -1 & 0 \\ 0 & 1
& 2&0 \end{array} \right) \rightarrow\ ({\rm row}\  2 \rightarrow
{\rm row}\  2 - {\rm row}\  1)$

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

$\left(\begin{array}{cccc} 1& 2 & 1 & 0\\ 0 & -1 & -2 & 0 \\ 0 & 0
& 0&0 \end{array} \right) $

Hence both $A$ and $A:b$ have rank 2
\item[(c)]
Hence equations do have a solution and since $r(A) = r(A:b)$

No of free parameters = no of unknowns $-\ r(A) = 3-2 = 1$

\item[(d)]
Equations are \begin{eqnarray*} x + 2y + z & = & 0 \\ -y - 2z & =
& 0 \end{eqnarray*}

Let $z = C \Rightarrow y = -2C \Rightarrow x = -2y - z = 4C - C =
3C$  and ${\bf x} = \left[\begin{array}{c} 3C \\ -2C \\ C
\end{array} \right]$ with one free variable as expected.
\end{description}


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