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%TCIDATA{Created=Mon Aug 19 14:52:24 1996}
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\begin{document}


\section{Equations\label{Systems of equations}}

Systems of equations were introduced in Chapter 2. The algebra of matrices
provides you with additional tools for solving systems of linear equations.

\subsection{\ Systems of Linear Equations}

You identify a system of equations by entering the equations in an $n\times
1 $ matrix, with one equation to a row. When you have the same number of
unknowns as equations, put the insertion point anywhere in the system, and
from the \textsf{Solve} submenu, choose \textsf{Exact}. The variables are
found automatically without having to be specified, as in the following
example. 
\index{Systems of equations@Systems of equations}

\begin{quote}
$\blacktriangleright $ \textsf{Solve + Exact}
\end{quote}

\begin{quotation}
$
\begin{array}{l}
x+y-2z=1\;\, \\ 
2x-4y+z=0\;\, \\ 
2y-3z=-1
\end{array}
$, Solution is : $\left\{ x=%
\frac{17}{8},y=\frac{11}{8},z=\frac{5}{4}\right\} $
\end{quotation}

To solve a system of equations with two equations and three unknowns, you
must specify \textsf{Variables to Solve for} in a dialog box. Put the
insertion point anywhere in the matrix and from the \textsf{Solve} submenu,
choose \textsf{Exact}. A dialog box comes up asking you to specify the
variables. Enter the variable names, separated by commas.

\begin{quote}
$\blacktriangleright $ \textsf{Solve + Exact}
\end{quote}

\begin{quotation}
\QTR{MenuDialog}{Variable(s) to Solve for} : $x,y$

$
\begin{array}{c}
2x-y=1 \\ 
x+3z=4
\end{array}
$, Solution is : $\left\{ y=-6z+7,x=-3z+4\right\} $

$\;$

\QTR{MenuDialog}{Variable(s) to Solve for} : $x,z$

$
\begin{array}{c}
2x-y=1 \\ 
x+3z=4
\end{array}
$, Solution is : $\left\{ x=\frac{1}{2}+\frac{1}{2}y,z=\frac{7}{6}-\frac{1}{6%
}y\right\} .\medskip $
\end{quotation}

\subsection{\ Matrix Equations\label{Matrix equations}}

The system of equations 
\begin{eqnarray*}
a_{11}x_{1}+a_{12}x_{2}+\ldots +a_{1n}x_{n} &=&b_{1} \\
a_{21}x_{1}+a_{22}x_{2}+\ldots +a_{2n}x_{n} &=&b_{2} \\
&&\vdots \\
a_{m1}x_{1}+a_{m2}x_{2}+\ldots +a_{mn}x_{n} &=&b_{m}
\end{eqnarray*}
is the same as the following matrix equation: 
\[
\left[ 
\begin{array}{cccc}
a_{11} & a_{12} & \ldots & a_{1n} \\ 
a_{21} & a_{22} & \ldots & a_{2n} \\ 
\vdots & \vdots & \ldots & \vdots \\ 
a_{m1} & a_{m2} & \ldots & a_{mn}
\end{array}
\right] \left[ 
\begin{array}{c}
x_{1} \\ 
x_{2} \\ 
\vdots \\ 
x_{n}
\end{array}
\right] =\left[ 
\begin{array}{c}
b_{1} \\ 
b_{2} \\ 
\vdots \\ 
b_{m}
\end{array}
\right] 
\]
You can solve these systems using \textsf{Exact} on the \textsf{Solve}
submenu. 
\index{Solve@Solve!exact@exact}\label{Solve, exact (3)}There are advantages
to solving systems of equations in this way, and often you can best deal
with systems of linear equations by solving the matrix version of the
system. Two of the preceding examples correspond to the first two of the
following matrix equations. Compare these results with the solutions
obtained previously.

\begin{example}
Multiply the coefficient matrix $\left[ 
\begin{array}{rrr}
1 & 1 & -2 \\ 
2 & -4 & 1 \\ 
0 & 2 & -3
\end{array}
\right] $ by the vector $\left[ 
\begin{array}{c}
x \\ 
y \\ 
z
\end{array}
\right] $ to display the system of equations $
\begin{array}{l}
x+y-2z=\;1 \\ 
2x-4y+z=\;0 \\ 
2y-3z=-1
\end{array}
$ in matrix form. 
\[
\left[ 
\begin{array}{rrr}
1 & 1 & -2 \\ 
2 & -4 & 1 \\ 
0 & 2 & -3
\end{array}
\right] \left[ 
\begin{array}{c}
x \\ 
y \\ 
z
\end{array}
\right] =\,\left[ 
\begin{array}{c}
x+y-2z \\ 
2x-4y+z \\ 
2y-3z
\end{array}
\right] =\left[ 
\begin{array}{r}
1 \\ 
0 \\ 
-1
\end{array}
\right] 
\]
\end{example}

\begin{quote}
$\blacktriangleright $ \textsf{Solve + Exact}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{rrr}
1 & 1 & -2 \\ 
2 & -4 & 1 \\ 
0 & 2 & -3
\end{array}
\right] \left[ 
\begin{array}{c}
x \\ 
y \\ 
z
\end{array}
\right] =\left[ 
\begin{array}{c}
1 \\ 
0 \\ 
-1
\end{array}
\right] $, Solution is : $\left[ 
\begin{array}{c}
\frac{17}{8}\vspace{6pt} \\ 
\vspace{6pt}\frac{11}{8} \\ 
\frac{5}{4}
\end{array}
\right] $

$\;$

$\left[ 
\begin{array}{rrr}
2 & -1 & 0 \\ 
1 & 0 & 3
\end{array}
\right] \left[ 
\begin{array}{c}
x \\ 
y \\ 
z
\end{array}
\right] =\left[ 
\begin{array}{c}
1 \\ 
4
\end{array}
\right] $, Solution is : $\left[ 
\begin{array}{c}
4-3t_{1} \\ 
7-6t_{1} \\ 
t_{1}
\end{array}
\right] $

$\;$

$\left[ 
\begin{array}{rrrc}
2 & -1 & 0 & 1 \\ 
1 & 0 & 3 & 1
\end{array}
\right] \left[ 
\begin{array}{c}
x \\ 
y \\ 
z \\ 
w
\end{array}
\right] =\left[ 
\begin{array}{c}
1 \\ 
4
\end{array}
\right] $, Solution is : $\left[ 
\begin{array}{c}
4-3t_{1}-t_{2} \\ 
7-6t_{1}-t_{2} \\ 
t_{1} \\ 
t_{2}
\end{array}
\right] $

\ 
\end{quotation}

In the first case, you can also solve the equation by multiplying both the
left and right sides of the equation by the inverse of the coefficient
matrix, and evaluating the product.

\begin{quote}
$\blacktriangleright $ \textsf{Evaluate}
\end{quote}

\begin{quotation}
$\left[ 
\begin{array}{c}
x \\ 
y \\ 
z
\end{array}
\right] =\left[ 
\begin{array}{rrr}
1 & 1 & -2 \\ 
2 & -4 & 1 \\ 
0 & 2 & -3
\end{array}
\right] ^{-1}\left[ 
\begin{array}{r}
1 \\ 
0 \\ 
-1
\end{array}
\right] =\,\left[ 
\begin{array}{c}
\vspace{6pt}\frac{17}{8} \\ 
\vspace{6pt}\frac{11}{8} \\ 
\frac{5}{4}\vspace*{6pt}
\end{array}
\right] \medskip $
\end{quotation}

\subsection{\ Related topics}

\begin{itemize}
\item  \hyperref{Working With Matrices}{}{}{DM8-1.tex#Matrix Algebra}

\item  \hyperref{Standard Operations}{}{}{DM8-2.tex#Standard matrix
operations}

\item  \hyperref{Row Operations and Echelon Forms}{}{}{DM8-3.tex#Matrices,
row operations}

\item  \hyperref{Matrix Operators}{}{}{DM8-5.tex#Trace}

\item  \hyperref{Polynomials and Vectors Associated with a Matrix}{}{}{%
DM8-6.tex#Characteristic, polynomial}

\item  \hyperref{Vector Spaces Associated with a Matrix}{}{}{DM8-7.tex#Row
space}

\item  \hyperref{Special Forms of Matrices}{}{}{DM8-8.tex#Smith normal form}

\item  \hyperref{Exercises and Solutions}{}{}{DM8-9.tex#Exercises}
\end{itemize}

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\end{document}
