% John David Stone
% Department of Mathematics and Computer Science
% Grinnell College
% stone@cs.grinnell.edu

% created June 19, 2002
% last revised June 19, 2002

\documentclass[11pt]{article}

\begin{document}
\textbf{Example 2A:} Solve the linear system $Ax = 0$, where $A$ is given
below, and find a basis for the solution set.  What is the dimension of the
solution set?
\begin{displaymath}
A := \left[ \begin{array}{rrrrr}
            -1 & -1 & -2 &  3 &  1 \\
            -9 &  5 & -4 & -1 & -5 \\
             7 & -5 &  2 &  3 &  5
            \end{array}                \right]
\end{displaymath}

\textbf{Solution:} We use \texttt{Matsolve} to solve $Ax = 0$ and then
decompose the solution to find a basis:

\begin{quote}
$> $ \texttt{Matsolve(A,zero,free=x);}
\end{quote}

The decomposition of this solution is

\begin{eqnarray*}
\left[ \begin{array}{c}
       -x_3 + x_4 \\  -x_3 + 2x_4 + x_5 \\ x_3 \\  x_4 \\ x_5
       \end{array} \right] & = &
  x_3 \left[ \begin{array}{c} -1 \\ -1 \\ 1 \\ 0 \\ 0 \end{array} \right] +
  x_4 \left[ \begin{array}{c}  1 \\  2 \\ 0 \\ 1 \\ 0 \end{array} \right] +
  x_5 \left[ \begin{array}{c}  0 \\  1 \\ 0 \\ 0 \\ 1 \end{array} \right]
\\
& = & x_3 u + x_4 v + x_5 w
\end{eqnarray*}
So the solution set is $\textrm{Span}\{u, v, w\}$.
\end{document}