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{\bf Question}

For the following system of equations $$ \left(\begin{array}{cccc}
1& 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\ 1 & 2 & 0 & 0 \\ 0 & 3 & -2 &
-2 \end{array} \right)\left(\begin{array}{c} x\\y\\z\\w\end{array}
\right) = \left(
\begin{array}{c} 0 \\ 4 \\ 1 \\ 1 \end{array} \right)$$
\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)]
$A = \left(\begin{array}{cccc} 1& 1 & 1 & 1 \\ 1 & 1 & -1 & -1 \\
1 & 2 & 0 & 0 \\ 0 & 3 & -2 & -2 \end{array} \right) \hspace{.3in}
A:b = \left(\begin{array}{ccccc} 1& 1 & 1 & 1 & 0\\ 1 & 1 & -1 &
-1 & 4 \\ 1 & 2 & 0 & 0 & 1 \\ 0 & 3 & -2 & -2 & 1 \end{array}
\right)$

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

$A:b = \left(\begin{array}{ccccc} 1& 1 & 1 & 1 & 0\\ 1 & 1 & -1 &
-1 & 4 \\ 1 & 2 & 0 & 0 & 1 \\ 0 & 3 & -2 & -2 & 1 \end{array}
\right) \rightarrow \begin{array}{l} \\ ({\rm row}\  2 \rightarrow
{\rm row}\  2 - {\rm row}\  1) \\ ({\rm row}\  3 \rightarrow {\rm
row}\  3 - {\rm row}\  1)
\\ \\
\end{array}$


$ = \left(\begin{array}{ccccc} 1& 1 & 1 & 1 & 0\\ 0 & 0 & -2 & -2
& 4 \\ 0 & 1 & -1 & -1 & 1 \\ 0 & 3 & -2 & -2 & 1 \end{array}
\right) \rightarrow \begin{array}{l}  ({\rm row}\  4 \rightarrow
{\rm row}\  4 - 3 {\rm row}\  3)
\end{array}$

$ = \left(\begin{array}{ccccc} 1& 1 & 1 & 1 & 0\\ 0 & 0 & -2 & -2
& 4 \\ 0 & 1 & -1 & -1 & 1 \\ 0 & 0 & 1 & 1 & -2 \end{array}
\right) \rightarrow \begin{array}{l} \\({\rm row}\  3 \rightarrow
{\rm row}\ 2)
\\ ({\rm row}\  2 \rightarrow {\rm row}\  4)
\\ ({\rm row}\  4 \rightarrow  {\rm row}\  3) \end{array}$

$ = \left(\begin{array}{ccccc} 1& 1 & 1 & 1 & 0 \\ 0 & 1 & -1 & -1
& 1 \\ 0 & 0 & 1 & 1 & -2 \\ 0 & 0 & -2 & -2 & 4 \end{array}
\right) \rightarrow \begin{array}{l} \\ ({\rm row}\  4 \rightarrow
{\rm row}\  4 + 2 {\rm row}\  3)
\end{array}$

$ = \left(\begin{array}{ccccc} 1& 1 & 1 & 1 & 0 \\ 0 & 1 & -1 & -1
& 1 \\ 0 & 0 & 1 & 1 & -2 \\ 0 & 0 & 0 & 0 & 0 \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 $-\ r(A) = 4-3 = 1$

\item[(d)]
Equations are \begin{eqnarray*} x + y + z + w & = & 0 \\ y - z - w
& = & 1 \\ z + w & = & -2\end{eqnarray*} Let $w = C \Rightarrow z
= -2 - C \Rightarrow y = -1 \Rightarrow x = 3$  and

${\bf x} = \left(\begin{array}{c} 3 \\ -1\\ -2-C \\ C
\end{array} \right)$ with one free variable as expected.
\end{description}



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