\hfill \thepage} %} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][P roof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Ma 116 Lecture 4/26/00} \vspace{1pt} \subsection{Matrices Continued} Last time we began discussing how to use matrices to solve systems of equations. We looked at an example and then talked about the elementary row operations that can be performed on a matrix. These are \vspace{1pt} \begin{enumerate} \item Interchanging the rows. \item Multiplying any row by a nonzero constant. \item Replacing any row by its sum with a nonzero constant multiple of any other row. (Add a multiple of one row to a different row.) \end{enumerate} \vspace{1pt} \paragraph{\protect\vspace{1pt}Example:} Find all solutions to the following system of equations \begin{eqnarray*} 3x+4y+z &=&1 \\ 2x+3y\text{ \ \ \ } &=&0 \\ 4x+3y-z &=&-2 \end{eqnarray*} We form what is called the \textit{augmented} matrix for this system. We now put this matrix in a form which will allow us to easily find the solution or soutions to the system. \begin{equation*} \left[ \begin{array}{cccc} 3 & 4 & 1 & 1 \\ 2 & 3 & 0 & 0 \\ 4 & 3 & -1 & -2% \end{array}% \right] \end{equation*} We could get a $1$ in the first row and first column by multiplying row $1$ $% \left( R_{1}\right) $ by $\frac{1}{3}.$ However, we can get a $1$ in this spot without obtaining fractions by subtracting row $2$ from row $1.$ Doing this we get \vspace{1pt} \begin{equation*} \left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 2 & 3 & 0 & 0 \\ 4 & 3 & -1 & -2% \end{array} \right] \end{equation*} We now use the $1$ that we have in the first row and first column to get zeroes below it. Hence we multiply row $1$ by $-2$ and add it to row $2.$We also multiply row $1$ by $-4$ and add it to row $3.$ we get \begin{equation*} \left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\ 0 & 1 & -2 & -2 \\ 0 & -1 & -5 & -6% \end{array} \right] \end{equation*} We now use the $1$ in the second row and second column to get zeroes above and below it. We get \vspace{1pt} \begin{equation*} \left[ \begin{array}{cccc} 1 & 0 & 3 & 3 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & -7 & -8% \end{array} \right] \end{equation*} \vspace{1pt} We may now get a $1$ in the third row and third column by multiplying the third row by $-\frac{1}{7}.$ Doing this yields \vspace{1pt} \begin{equation*} \left[ \begin{array}{cccc} 1 & 0 & 3 & 3 \\ 0 & 1 & -2 & -2 \\ 0 & 0 & 1 & \frac{8}{7}% \end{array} \right] \end{equation*} \vspace{1pt} Now we can use the one in the last row to get zeroes for the entries above it. Doing this we get \vspace{1pt} \begin{equation*} \left[ \begin{array}{cccc} 1 & 0 & 0 & -\frac{3}{7} \\ 0 & 1 & 0 & \frac{2}{7} \\ 0 & 0 & 1 & \frac{8}{7}% \end{array} \right] \end{equation*} \vspace{1pt} Clearly the solution is $x=-\frac{3}{7},y=\frac{2}{7},z=\frac{8}{7}.$ \paragraph{\protect\vspace{1pt}Example:} \vspace{1pt} Solve the system \begin{eqnarray*} x_{1}-3x_{2}+x_{3}-x_{4} &=&-1 \\ -x_{1}+3x_{2}+3x_{4}+x_{5} &=&3 \\ 2x_{1}-6x_{2}+3x_{3}-x_{5} &=&2 \\ -x_{1}+3x_{2}+x_{3}+5x_{4}+x_{5} &=&6 \end{eqnarray*} \vspace{1pt}using the Gauss algorithm and back substitution. The augmented matrix is \begin{center} $\left[ \begin{array}{cccccc} 1 & -3 & 1 & -1 & 0 & -1 \\ -1 & 3 & 0 & 3 & 1 & 3 \\ 2 & -6 & 3 & 0 & -1 & 2 \\ -1 & 3 & 1 & 5 & 1 & 6% \end{array} \right] $, row echelon form: $\left[ \begin{array}{cccccc} 1 & -3 & 0 & -3 & 0 & -4 \\ 0 & 0 & 1 & 2 & 0 & 3 \\ 0 & 0 & 0 & 0 & 1 & -1 \\ 0 & 0 & 0 & 0 & 0 & 0% \end{array} \right] $ \vspace{1pt} \end{center} The corresponding equations are \vspace{1pt} \begin{eqnarray*} x_{1}-3x_{2}-3x_{4} &=&-4 \\ x_{3}+2x_{4} &=&3 \\ x_{5} &=&-1 \end{eqnarray*} \vspace{1pt} Thus \vspace{1pt} \begin{center} $x_{5}=-1,\qquad x_{3}=3-2x_{4},\qquad x_{1}=3x_{2}+3x_{4}-4$ \vspace{1pt} \end{center} Letting $x_{2}=s$ and $x_{4}=t$ we have the infinite set of solutions \vspace{1pt} \begin{center} $x_{1}=3s+3t-4,\qquad x_{3}=3-2t,\qquad x_{5}=-1$ \vspace{1pt} \end{center} \subsection{Inverse of a Matrix} \vspace{1pt}Definition: If $A$ is a square $n\times n$, a matrix $B$ is called the \emph{inverse} of $A$ if and only if \vspace{1pt} \begin{center} $AB=I$ \ \ and $\ \ \ BA=I.$ \vspace{1pt} \end{center} \vspace{1pt}A matrix $A$ that has an inverse is called an \emph{invertible or nonsingular matrix.} \paragraph{Example} \vspace{1pt}Show that the matrix $B=\left[ \begin{array}{cc} -1 & 1 \\ 1 & 0% \end{array} \right] $ is an inverse of $A=\left[ \begin{array}{cc} 0 & 1 \\ 1 & 1% \end{array} \right] .$ \vspace{1pt} $BA=\left[ \begin{array}{cc} -1 & 1 \\ 1 & 0% \end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 1 & 1% \end{array} \right] =\allowbreak \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1% \end{array} \right] $ and $AB=\left[ \begin{array}{cc} 0 & 1 \\ 1 & 1% \end{array} \right] \left[ \begin{array}{cc} -1 & 1 \\ 1 & 0% \end{array} \right] =\allowbreak \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1% \end{array} \right] ,$ so $B$ is indeed an inverse of $A.$ \vspace{1pt} \paragraph{Example} The matrix $A=\left[ \begin{array}{cc} 0 & 0 \\ 1 & 1% \end{array} \right] $ is not invertible. For if $B=\left[ \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22}% \end{array} \right] $ is any $2\times 2$ matrix, then \vspace{1pt} $\left[ \begin{array}{cc} 0 & 0 \\ 1 & 1% \end{array} \right] \left[ \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22}% \end{array} \right] =\allowbreak \left[ \begin{array}{cc} 0 & 0 \\ b_{11}+b_{21} & b_{12}+b_{22}% \end{array} \right] \neq \left[ \begin{array}{cc} 1 & 0 \\ 0 & 1% \end{array} \right] =I.$ \vspace{1pt} \paragraph{Theorem} If $B$ and $C$ are both inverses of $A,$ then $B=C.$ \vspace{1pt} Remark: If $A$ is invertible then the (unique) inverse of $A$ is denoted by $% A^{-1}.$ \paragraph{Example:} \vspace{1pt}Find $A^{-1}$ for \begin{equation*} A=\left[ \begin{array}{cc} 2 & 3 \\ 1 & 4% \end{array}% \right] . \end{equation*} We seek a matrix \begin{equation*} A^{-1}=\left[ \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22}% \end{array}% \right] \end{equation*} such that \vspace{1pt} \begin{equation*} \left[ \begin{array}{cc} 2 & 3 \\ 1 & 4% \end{array}% \right] \left[ \begin{array}{c} b_{11} \\ b_{21}% \end{array}% \right] =\left[ \begin{array}{c} 1 \\ 0% \end{array}% \right] \end{equation*} \ and \vspace{1pt} \vspace{1pt} \begin{equation*} \left[ \begin{array}{cc} 2 & 3 \\ 1 & 4% \end{array}% \right] \left[ \begin{array}{c} b_{12} \\ b_{22}% \end{array}% \right] =\left[ \begin{array}{c} 0 \\ 1% \end{array}% \right] \end{equation*} \vspace{1pt} We could form two augmented matrices (one for each system) and then put each of them in what is called reduced row-echelon form. That is, we want to get the matrix $\left[ \begin{array}{cc} 2 & 3 \\ 1 & 4% \end{array}% \right] $ in the form $\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1% \end{array}% \right] $ while keeping track of what happens to the third column of the augmented matrix. However, we may just as well do the entire reduction at the same time. \vspace{1pt} \begin{equation*} \left[ \begin{array}{llll} 2 & 3 & 1 & 0 \\ 1 & 4 & 0 & 1% \end{array}% \right] \longrightarrow \left[ \begin{array}{llll} 1 & 4 & 0 & 1 \\ 2 & 3 & 1 & 0% \end{array}% \right] \longrightarrow \left[ \begin{array}{llll} 1 & 4 & 0 & 1 \\ 0 & -5 & 1 & -2% \end{array}% \right] \end{equation*} \vspace{1pt} \begin{equation*} \longrightarrow \left[ \begin{array}{llll} 1 & 4 & 0 & 1 \\ 0 & 1 & -\frac{1}{5} & \frac{2}{5}% \end{array}% \right] \longrightarrow \left[ \begin{array}{llll} 1 & 0 & +\frac{4}{5} & -\frac{3}{5} \\ 0 & 1 & -\frac{1}{5} & \frac{2}{5}% \end{array}% \right] \end{equation*}% $\qquad $% \begin{equation*} A^{-1}=\left[ \begin{array}{cc} \frac{4}{5} & -\frac{3}{5} \\ -\frac{1}{5} & \frac{2}{5}% \end{array}% \right] \end{equation*} \vspace{1pt} \paragraph{Example: (not to be presented in class)} Under what conditions is the $2\times 2$ matrix $A=\left[ \begin{array}{cc} a & b \\ c & d% \end{array}% \right] $ $\left( a\neq 0\right) $ invertible. When $A$ is invertible, find $% A^{-1}.$ \vspace{1pt} We seek a matrix $A^{-1}=\left[ \begin{array}{cc} b_{11} & b_{12} \\ b_{21} & b_{22}% \end{array} \right] $ such that \vspace{1pt} $\left[ \begin{array}{cc} a & b \\ c & d% \end{array}% \right] \left[ \begin{array}{c} b_{11} \\ b_{21}% \end{array}% \right] =\left[ \begin{array}{c} 1 \\ 0% \end{array}% \right] $ \ and \vspace{1pt} \vspace{1pt} $\left[ \begin{array}{cc} a & b \\ c & d% \end{array}% \right] \left[ \begin{array}{c} b_{12} \\ b_{22}% \end{array}% \right] =\left[ \begin{array}{c} 0 \\ 1% \end{array}% \right] $ \vspace{1pt} We could form two augmented matrices (one for each system) and then put each of them in what is called reduced row-echelon form. That is, we want to get the matrix $\left[ \begin{array}{cc} a & b \\ c & d% \end{array}% \right] $ in the form $\left[ \begin{array}{cc} 1 & 0 \\ 0 & 1% \end{array}% \right] $ while keeping track of what happens to the third column of the augmented matrix. However, we may just as well do the entire reduction at the same time. Thus \vspace{1pt} $\left[ \begin{array}{cccc} a & b & 1 & 0 \\ c & d & 0 & 1% \end{array}% \right] \rightarrow \left[ \begin{array}{cccc} 1 & \dfrac{b}{a} & \dfrac{1}{a} & 0 \\ c & d & 0 & 1% \end{array}% \right] \rightarrow $ $\left[ \begin{array}{cccc} 1 & \dfrac{b}{a} & \dfrac{1}{a} & 0 \\ 0 & \dfrac{da-bc}{a} & -\dfrac{c}{a} & 1% \end{array}% \right] \rightarrow \left[ \begin{array}{cccc} 1 & \dfrac{b}{a} & \dfrac{1}{a} & 0 \\ 0 & \dfrac{da-bc}{a} & -\dfrac{c}{a} & 1% \end{array}% \right] $ $\left[ \begin{array}{cccc} 1 & 0 & \frac{d}{da-cb} & -\frac{b}{da-cb} \\ 0 & 1 & -\frac{c}{da-cb} & \frac{1}{da-cb}a% \end{array}% \right] .$ \vspace{1pt} Thus we see that $A$ is invertible $\Longleftrightarrow $ $ad-bc\neq 0.$ If this condition holds, then \vspace{1pt} \begin{center} $A^{-1}=\dfrac{1}{ad-bc}\left[ \begin{array}{cc} d & -b \\ -c & a% \end{array}% \right] $ \vspace{1pt} \end{center} \paragraph{Example:} Find $A^{-1}$ for $A=\left[ \begin{array}{ccc} 2 & 7 & 1 \\ 1 & 4 & -1 \\ 1 & 3 & 0% \end{array} \right] $. We form $\left[ \begin{array}{cccccc} 2 & 7 & 1 & 1 & 0 & 0 \\ 1 & 4 & -1 & 0 & 1 & 0 \\ 1 & 3 & 0 & 0 & 0 & 1% \end{array} \right] .$ $\vspace{1pt}$ $\vspace{1pt}\left[ \begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1% \end{array} \right] \left[ \begin{array}{cccccc} 2 & 7 & 1 & 1 & 0 & 0 \\ 1 & 4 & -1 & 0 & 1 & 0 \\ 1 & 3 & 0 & 0 & 0 & 1% \end{array} \right] =\allowbreak \left[ \begin{array}{cccccc} 1 & 4 & -1 & 0 & 1 & 0 \\ 2 & 7 & 1 & 1 & 0 & 0 \\ 1 & 3 & 0 & 0 & 0 & 1% \end{array} \right] $ \vspace{1pt} \section{Determinants} \vspace{1pt} \begin{center} \vspace{1pt} \end{center} \subsection{Determinants of 2x2 and 3x3 Matrices} With each \emph{square} matrix we can associate a number called the \emph{% determinant} of the matrix. Since the actual definition of a determinant is somewhat involved, we begin by showing how one evaluates $2\times 2$ and $% 3\times 3$ matrices. The symbols $\left| {}\right| $ and $\det $ are used to denote the determinant of a matrix. For $2\times 2$ matrices we have \begin{equation*} \det \left[ \begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22}% \end{array}% \right] =\left| \begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22}% \end{array}% \right| =a_{11}a_{22}-a_{12}a_{21} \end{equation*} \paragraph{Example:\qquad \qquad \qquad} $\left| \begin{array}{ll} 1 & -5 \\ 2 & -1% \end{array} \right| =1\left( -1\right) -\left( -5\right) \left( 2\right) =-1+10=9$ \vspace{1pt} For $3\times 3$ matrices we have \begin{eqnarray*} \det \left[ \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}% \end{array}% \right] &=&\left| \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}% \end{array}% \right| \\ &=&+a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{31}a_{22}a_{13}-a_{32}a_{23}a_{11}-a_{33}a_{21}a_{12} \end{eqnarray*} $\qquad \qquad \qquad $ \paragraph{Example:} $\left| \begin{array}{ccc} 1 & 2 & -1 \\ 0 & -2 & 1 \\ 3 & 1 & -2% \end{array} \right| =\left( 1\right) \left( -2\right) \left( -2\right) +\left( 2\right) \left( 1\right) \left( 3\right) +\left( -1\right) \left( 0\right) \left( 1\right) -\left( 3\right) \left( -2\right) \left( -1\right) -\left( 1\right) \left( 1\right) \left( 1\right) -\left( -2\right) \left( 0\right) \left( 2\right) =3$ \vspace{1pt} Note that SNB will evaluate matrices of any order. Simply click Maple, Matrices, Determinant, or put the curser in the determinant and evaluate. \vspace{1pt} \paragraph{Example:} \begin{equation*} \left| \begin{array}{cccc} 10 & -6 & 4 & 9 \\ -2 & 3 & 0 & 6 \\ 1 & -2 & 5 & 10 \\ -6 & 9 & 0 & 3% \end{array}% \right| =\allowbreak -1410 \end{equation*} \vspace{1pt} \subsection{The Definition of the Determinant of a Square Matrix} \vspace{1pt}In all that follows $A$ is a square $n\times n$ matrix. \vspace{1pt} We shall need some facts about permutations to define the determinant of a square matrix. \subsubsection{Permutations:} \vspace{1pt} Definition: A rearrangement of symbols is called a permutation. \vspace{1pt} Ex. $1,2,3$ \begin{equation*} 123\text{ \ \ }231\text{ \ \ }132\text{ \ \ }213\text{ \ \ }312\text{ \ \ }% 321 \end{equation*} $\ $ \vspace{1pt} There are $3!=6$ permutations Given the integers $1,2,..,n$ there are $n!$ permutations possible. Definition. If in a given permutation a larger integer precedes a smaller one, we say that there is an \emph{inversion}. If in a given permutation the number of inversions is \emph{even} (\emph{odd}% ), the permutation is called \emph{even} (\emph{odd}). \vspace{1pt} Ex. $123$ no inversions - even \qquad\ \qquad\ $312$ $\ \ \ \ \ \ 2$ inversions: $3$ precedes $1,$ $\ 3$ precedes $% 2 $ $\Longrightarrow $even \qquad\ \qquad\ $4213$ $\ \ \ \ 3$ inversions for the $4$, $1$ inversion for the $2,$ $\qquad \qquad \qquad \qquad \qquad 3+1$ or $4$ inversions \ - hence even \vspace{1pt} \qquad\ $2431$ $\ \ \ 1,$ $2,$ $1\Longrightarrow 4$ even \vspace{1pt} \subsection{The Determinant of a Square Matrix} \vspace{1pt} Consider \begin{equation*} A=\left[ a_{ij}\right] _{n\times n}=\left[ \begin{array}{llll} a_{11} & a_{12} & . & a_{1n} \\ . & . & . & . \\ . & . & . & . \\ a_{n1} & . & . & a_{nn}% \end{array}% \right] _{n\times n} \end{equation*} \vspace{1pt} and a product $\qquad \qquad $ $\vspace{1pt}$ \begin{equation} a_{1j_{1}}a_{2j_{2}}a._{3j_{3}}\cdots a_{nj_{n}} \tag{$\left( 1\right) $} \end{equation} \qquad of $n$ of its elements selected so that one and only one element comes from any row and one and only one element comes from any column. In $(1)$ the first subscripts are arranged in the order $1,2,.$ $.$ $.$ $,n$. The sequence of subscripts $j_{1},.$ $.$ $.$ $,j_{n}$ is one of the $n!$ permutations of $1,2,$ $.$ $.$ $.$ $,n$. \vspace{1pt} For a given permutation $j_{1},j_{2},$ $.$ $.$ $.$ $,j_{n}$ define \vspace{1pt} \begin{center} \vspace{1pt} \begin{equation*} \epsilon _{j_{1}j_{2}.\text{ }.\text{ }.\text{ }j_{n}}=\left\{ \begin{array}{c} +1\qquad if\text{ }permutation\text{ }is\text{ }even \\ -1\text{ \ \ \ \ }if\text{ }permutation\text{ }is\text{ }odd% \end{array} \right. \end{equation*} \end{center} \vspace{1pt} and consider \begin{center} \qquad \qquad \begin{equation} \epsilon _{j_{1}j_{2}.\text{ }.\text{ }.\text{ }j_{n}}a_{1j_{1}}a_{2j_{2}}% \cdots a_{nj_{n}} \tag{$\left( 2\right) $} \end{equation} $\qquad $ \end{center} \vspace{1pt} By the determinant of $A$ of order $n$, denoted by $|A|$ or $\det A$ we mean the sum of all the different signed products of the form $(2)$. Thus \begin{center} \qquad \qquad \begin{equation*} |A|=\sum\limits_{\rho }\epsilon _{j_{1}j_{2}.\text{ }.\text{ }.\text{ }% j_{n}}a_{1j_{1}}a_{2j_{2}}\cdots a_{nj_{n}} \end{equation*} \end{center} \vspace{1pt}where $\rho $ ranges over the $n!$ permutations of $1,2,$ $.\,.$ $.$ $,n$. \vspace{1pt} \paragraph{Example:} $\left| \begin{array}{ll} a_{11} & a_{12} \\ a_{21} & a_{22}% \end{array} \right| =\epsilon _{12}a_{11}a_{22}+\epsilon _{21}a_{12}a_{21}=a_{11}a_{22}-a_{12}a_{21}$ \qquad \qquad \qquad \qquad\ even\qquad\ \ \ odd \qquad \qquad \paragraph{\protect\vspace{1pt}Example:} \begin{equation*} \left| \begin{array}{ll} 1 & -5 \\ 2 & -1% \end{array}% \right| =-1+10=9 \end{equation*} \vspace{1pt} \vspace{1pt}\vspace{1pt} \end{document} %%%%%%%%%%%%%%%%%%%% End /document/lec_4_24_00.tex %%%%%%%%%%%%%%%%%%%%