\hfill \thepage} %} \input{tcilatex} \begin{document} \section{Ma 112 Lecture 2/4/99} \subsection{Solving Linear Equations: Chapter 2{}} \vspace{1pt} \subsection{Elimination Using Matrices: 2.3} \vspace{1pt} \paragraph{Example:} \vspace{1pt} Solve the system \begin{eqnarray*} x_{1}+x_{2}+2x_{3}+x_{4} &=&5 \\ 2x_{1}+3x_{2}-x_{3}-2x_{4} &=&2 \\ 4x_{1}+5x_{2}+2x_{3} &=&7 \end{eqnarray*} \vspace{1pt} \begin{center} $ \begin{array}{c} x_{1}+x_{2}+2x_{3}+x_{4}=5 \\ 2x_{1}+3x_{2}-x_{3}-2x_{4}=2 \\ 4x_{1}+5x_{2}+2x_{3}=7 \end{array} \qquad \longleftrightarrow \qquad \left[ \begin{array}{lllll} 1 & 1 & 2 & 1 & 5 \\ 2 & 3 & -1 & -2 & 2 \\ 4 & 5 & 2 & 0 & 7 \end{array} \right] $ \vspace{1pt} \end{center} The matrix on the right that we have associated with the given system is called the \emph{augmented matrix }of the system.\emph{\ }The matrix \vspace{1pt} \begin{center} $A=\left[ \begin{array}{llll} 1 & 1 & 2 & 1 \\ 2 & 3 & -1 & -2 \\ 4 & 5 & 2 & 0 \end{array} \right] $ \end{center} \vspace{1pt}is called the \emph{coefficient matrix }of the system, and $C=% \left[ \begin{array}{l} 5 \\ 2 \\ 7 \end{array} \right] $ is called the \emph{constant matrix (vector)} of the system. It is clear that we can rewrite our system as \[ AX=C \] \vspace{1pt} where $X=\left[ \begin{array}{c} x_{1} \\ x_{2} \\ x_{3} \\ x_{4} \end{array} \right] .$ \begin{center} \vspace{1pt} $ \begin{array}{c} x_{1}+x_{2}+2x_{3}+x_{4}=5 \\ 0+x_{2}-5x_{3}-4x_{4}=-8 \\ 0+x_{2}-6x_{3}-4x_{4}=-13 \end{array} \longleftrightarrow \left[ \begin{array}{lllll} 1 & 1 & 2 & 1 & 5 \\ 0 & 1 & -5 & -4 & -8 \\ 0 & 1 & -6 & -4 & -13 \end{array} \right] $ \end{center} $\qquad \qquad $ \begin{center} $ \begin{array}{c} x_{1}+0x_{2}+7x_{3}+5x_{4}=13 \\ 0+x_{2}-5x_{3}-4x_{4}=-8 \\ 0+0-x_{3}=-5 \end{array} \longleftrightarrow \left[ \begin{array}{lllll} 1 & 0 & 7 & 5 & 13 \\ 0 & 1 & -5 & -4 & -8 \\ 0 & 0 & -1 & 0 & -5 \end{array} \right] $ \end{center} \vspace{1pt} \begin{center} \qquad $ \begin{array}{c} x_{1}+0x_{2}+7x_{3}+5x_{4}=13 \\ 0+x_{2}-5x_{3}-4x_{4}=-8 \\ 0+0+x_{3}=5 \end{array} \longleftrightarrow \left[ \begin{array}{lllll} 1 & 0 & 0 & 5 & -22 \\ 0 & 1 & 0 & -4 & 17 \\ 0 & 0 & 1 & 0 & 5 \end{array} \right] $ \end{center} $\qquad $ Thus $x_{3}=5,\qquad x_{1}+5x_{4}=-22,\qquad x_{2}-4x_{4}=17$ or $x_{3}=5,\qquad x_{4}=t\qquad x_{1}=-5t-22\qquad x_{2}=4t+17$ \vspace{1pt} Note that we have an infinite number of solutions. However, if our operations had led to \begin{center} \vspace{1pt}$\left[ \begin{array}{lllll} 1 & 0 & 0 & 5 & -22 \\ 0 & 1 & 0 & -4 & 17 \\ 0 & 0 & 0 & 0 & 5 \end{array} \right] ,$ \vspace{1pt} \end{center} then there would not be any solution to the system, since the last row of the matrix would imply \vspace{1pt} \[ 0x_{1}+0x_{2}+0x_{3}+0x_{4}=5 \] \vspace{1pt} which is clearly impossible. \vspace{1pt} Definition: Systems of linear equations that have no solution are called \emph{inconsistent systems}; systems that have at least one solution are said to be \emph{consistent.} \vspace{1pt} \subsubsection{SNB ICE} Find a condition on the numbers $a,b,$ and $c$ such that the following system of equations is consistent. When that condition is satisfied, find all solutions (in terms of $a,b,$ and $c).$ Show all steps in your work. \vspace{1pt} \begin{eqnarray*} x+3y+z &=&a \\ -x-2y+z &=&b \\ 3x+7y-z &=&c \end{eqnarray*} \vspace{1pt} \paragraph{Example:} Note that the operations we have done on the matrix above are equivalent to multiplying the matrix by a ``modified'' $3\times 3$ identity matrix. Thus \vspace{1pt} $\left[ \begin{array}{ccc} 1 & 0 & 0 \\ -2 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{lllll} 1 & 1 & 2 & 1 & 5 \\ 2 & 3 & -1 & -2 & 2 \\ 4 & 5 & 2 & 0 & 7 \end{array} \right] =\allowbreak \left[ \begin{array}{ccccc} 1 & 1 & 2 & 1 & 5 \\ 0 & 1 & -5 & -4 & -8 \\ 4 & 5 & 2 & 0 & 7 \end{array} \right] $ \vspace{1pt} $\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -4 & 0 & 1 \end{array} \right] \left[ \begin{array}{ccccc} 1 & 1 & 2 & 1 & 5 \\ 0 & 1 & -5 & -4 & -8 \\ 4 & 5 & 2 & 0 & 7 \end{array} \right] =\allowbreak \left[ \begin{array}{ccccc} 1 & 1 & 2 & 1 & 5 \\ 0 & 1 & -5 & -4 & -8 \\ 0 & 1 & -6 & -4 & -13 \end{array} \right] $ \vspace{1pt} Note also that $\left[ \begin{array}{ccc} 0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0 \end{array} \right] \left[ \begin{array}{lllll} 1 & 1 & 2 & 1 & 5 \\ 2 & 3 & -1 & -2 & 2 \\ 4 & 5 & 2 & 0 & 7 \end{array} \right] =\allowbreak \left[ \begin{array}{ccccc} 4 & 5 & 2 & 0 & 7 \\ 2 & 3 & -1 & -2 & 2 \\ 1 & 1 & 2 & 1 & 5 \end{array} \right] ,$ \vspace{1pt} whereas as \vspace{1pt} $\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & 1 \end{array} \right] \left[ \begin{array}{lllll} 1 & 1 & 2 & 1 & 5 \\ 2 & 3 & -1 & -2 & 2 \\ 4 & 5 & 2 & 0 & 7 \end{array} \right] =\allowbreak \left[ \begin{array}{ccccc} 1 & 1 & 2 & 1 & 5 \\ -4 & -6 & 2 & 4 & -4 \\ 4 & 5 & 2 & 0 & 7 \end{array} \right] $ \vspace{1pt} Thus we see that by multiplying the given matrix by various modifications of the identity matrix, we can make certain elements in the matrix a $0$, interchange two rows, and multiply a given row by a scalar. \vspace{1pt} \end{document} %%%%%%%%%%%%%%%%%%% End /document/lec2_4_99_done.tex %%%%%%%%%%%%%%%%%%