\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][Proof]{\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \begin{document} \section{Ma 116 Lecture 4/24/00} \subsection{Multiplication of Matrices} \qquad \qquad Consider a system of $m$ equations in $n$ unknowns \qquad \qquad \begin{center} \begin{eqnarray*} a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{1n}x_{n} &=&d_{1} \\ a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n} &=&d_{2} \\ \cdots &=&\cdots \qquad \\ \cdots &=&\cdots \\ \ddots &=&\ddots \\ a_{m1}x_{1}+a_{m2}x_{2}+\cdots +a_{mn}x_{n} &=&d_{m} \end{eqnarray*} \vspace{1pt} \end{center} Here the $a_{ij}$ and $d_{i}$ are given scalars and the $x_{j}$ are the unknowns. \begin{center} \qquad \qquad \qquad \qquad \end{center} If we let $A=\left[ a_{ij}\right] _{m\times n}$\ , \vspace{1pt} \begin{center} $X=\left[ \begin{array}{l} x_{1} \\ \vdots \\ \vdots \\ x_{n}% \end{array} \right] _{n\times 1},\qquad $and$\qquad D=\left[ \begin{array}{l} d_{1} \\ \vdots \\ \vdots \\ d_{m}% \end{array} \right] _{m\times 1}$ \end{center} \vspace{1pt} \qquad \qquad \qquad \qquad $\qquad \qquad \qquad \qquad \qquad $ Then it is natural to write $AX=D$ to represent the system above. Hence we want \vspace{1pt} \begin{center} $\left[ \begin{array}{lllll} a_{11} & \cdots & \cdots & \cdots & a_{1m} \\ a_{21} & \cdots & \cdots & \cdots & a_{2m} \\ \cdots & \cdots & \cdots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{m1} & \cdots & \cdots & \cdots & a_{mn}% \end{array} \right] \left[ \begin{array}{l} x_{1} \\ x_{2} \\ \vdots \\ \vdots \\ x_{n}% \end{array} \right] =\left[ \begin{array}{l} d_{1} \\ d_{2} \\ \vdots \\ \vdots \\ d_{m}% \end{array} \right] $ \end{center} \vspace{1pt} to be the same as our system above. $\Longrightarrow $ that multiplication should be defined by \vspace{1pt} \begin{center} \qquad \qquad $d_{i}=\sum\limits_{j=1}^{n}a_{ij}x_{j}$ \end{center} Notice that $A$ is $m\times n$ and $X$ is $n\times 1$ and $D$ is $m\times 1$% . Thus to multiply two matrices we must have the number of columns of the first matrix equal to the number of rows of the second matrix. To multiply two matrices $A$ and $B$ together, where $B$ is not a column matrix, we extrapolate as follows: \vspace{1pt} \vspace{1pt} \begin{center} $\left[ \begin{array}{lllll} a_{11} & \cdots & \cdots & \cdots & a_{1m} \\ a_{21} & \cdots & \cdots & \cdots & a_{2m} \\ \cdots & \cdots & \cdots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a_{m1} & \cdots & \cdots & \cdots & a_{mn}% \end{array} \right] _{m\times n}\left[ \begin{array}{lllll} b_{11} & \cdots & \cdots & \cdots & b_{1p} \\ b_{21} & \cdots & \cdots & \cdots & b_{2p} \\ \cdots & \cdots & \cdots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ b_{n1} & \cdots & \cdots & \cdots & b_{np}% \end{array} \right] _{n\times p}=\left[ \begin{array}{lllll} c_{11} & \cdots & \cdots & \cdots & c_{1p} \\ c_{21} & \cdots & \cdots & \cdots & c_{2p} \\ \cdots & \cdots & \cdots & \ddots & \vdots \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ c_{m1} & \cdots & \cdots & \cdots & c_{mp}% \end{array} \right] _{m\times p}$ \vspace{1pt} \end{center} We see that \vspace{1pt} \begin{equation*} c_{11}=\sum\limits_{k=1}^{n}a_{1k}b_{k1} \end{equation*} \vspace{1pt}and in general that \begin{center} \begin{equation*} c_{ij}=\sum\limits_{k=1}^{n}a_{ik}b_{kj} \end{equation*} \end{center} \vspace{1pt} Definition. Let $A=\left[ a_{ij}\right] _{m\times n}$ and $B=\left[ b_{ij}% \right] _{n\times p}$ be matrices. Then $A$ $B$ is the $m\times p$ matrix $% C, $ where \vspace{1pt} \begin{center} \vspace{1pt}\qquad \qquad \begin{equation*} C=\left[ c_{ij}\right] _{m\times p}=\left[ \sum\limits_{k=1}^{n}a_{ik}b_{kj}% \right] _{m\times p} \end{equation*} \end{center} \vspace{1pt} Remark. $A$ $B\neq B$ $\,A$ necessarily. \vspace{1pt} In fact $B$ $A$ need not be defined. For example if $A$ is $2\times 3$ and \ $B$ is $3\times 4$, then $A$ $B$ will be $2\times 4$ whereas $B$ $A$ is not defined. \vspace{1pt} \vspace{1pt} \paragraph{Example:} $\left[ \begin{array}{lll} 1 & -1 & 0 \\ 4 & 1 & -1% \end{array} \right] _{2\times 3}\times \left[ \begin{array}{ll} 3 & 4 \\ -1 & -5 \\ 1 & 2% \end{array} \right] _{3\times 2}=\left[ \begin{array}{cc} \left( 1\right) \left( 3\right) +(-1)\left( -1\right) +\left( 0\right) \left( 1\right) & \left( 1\right) \left( 4\right) +\left( -1\right) \left( -5\right) +\left( 0\right) \left( 2\right) \\ \left( 4\right) \left( 3\right) +\left( 1\right) \left( -1\right) +\left( -1\right) \left( 1\right) & \left( 4\right) \left( 4\right) +\left( 1\right) \left( -5\right) +\left( -1\right) \left( 2\right)% \end{array} \right] _{2\times 2}$ $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad =% \left[ \begin{array}{ll} 4 & 9 \\ 10 & 9% \end{array} \right] _{2\times 2}$ \vspace{1pt} \qquad \qquad \qquad $\qquad \qquad \qquad $ \vspace{1pt} $\left[ \begin{array}{ll} 3 & 4 \\ -1 & -5 \\ 1 & 2% \end{array} \right] _{3\times 2}\times \left[ \begin{array}{lll} 1 & -1 & 0 \\ 4 & 1 & -1% \end{array} \right] _{2\times 3}=\left[ \begin{array}{lll} 19 & 1 & -4 \\ -21 & -4 & 5 \\ 9 & 1 & -2% \end{array} \right] _{3\times 3}$ \vspace{1pt} Note that using Evaluate in SNB gives the same result. \begin{center} \vspace{1pt} $\left[ \begin{array}{ll} 3 & 4 \\ -1 & -5 \\ 1 & 2% \end{array} \right] \left[ \begin{array}{lll} 1 & -1 & 0 \\ 4 & 1 & -1% \end{array} \right] =\allowbreak \left[ \begin{array}{ccc} 19 & 1 & -4 \\ -21 & -4 & 5 \\ 9 & 1 & -2% \end{array} \right] $ \end{center} \qquad \qquad \qquad $\qquad \qquad \qquad \qquad $ The following occur often for matrices. \vspace{1pt} \begin{enumerate} \item $A$ $B\neq B$ $A$ \end{enumerate} \paragraph{Example:} $\left[ \begin{array}{cc} 1 & 0 \\ 0 & 0% \end{array} \right] \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0% \end{array} \right] =\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0% \end{array} \right] \neq \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0% \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0% \end{array} \right] $ $=$ $\left[ \begin{array}{cc} 0 & 0 \\ 0 & 0% \end{array} \right] $ \begin{enumerate} \item[2.] $A$ $B=0$ but neither $A=0$ or $B=0$ \end{enumerate} \paragraph{Example:} $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0% \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0% \end{array} \right] $ $=$ $\left[ \begin{array}{cc} 0 & 0 \\ 0 & 0% \end{array} \right] $ \begin{enumerate} \item[3.] $A$ $B=A$ $C$\ \ but $B\neq C$ \end{enumerate} \vspace{1pt} $\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0% \end{array} \right] \left[ \begin{array}{cc} 1 & 0 \\ 0 & 0% \end{array} \right] =\left[ \begin{array}{cc} 0 & 1 \\ 0 & 0% \end{array} \right] \left[ \begin{array}{cc} 5 & 0 \\ 0 & 0% \end{array} \right] =\left[ \begin{array}{cc} 0 & 0 \\ 0 & 0% \end{array} \right] $ \vspace{1pt} \paragraph{Theorem} \vspace{1pt}Assume that $k$ is an arbitrary scalar, and that $A,$ $B,$ $C$ and $I$ are matrices of sizes such that the indicated operations can be performed. Then \vspace{1pt} $1.$ $IA=A,\qquad BI=B$ \vspace{1pt} $2.$ $A\left( BC\right) =\left( AB\right) C$ \vspace{1pt} $3.$ \ $A\left( B+C\right) =AB+AC,\qquad A\left( B-C\right) =AB-AC$ \vspace{1pt} $4.$ \ $\left( B+C\right) A=BA+CA,\qquad \left( B-C\right) A=BA-CA$ \vspace{1pt} $5.$ \ $k\left( AB\right) =\left( kA\right) B=A\left( kB\right) $ \vspace{1pt} $6.$ \ $\left( AB\right) ^{T}=B^{T}A^{T}.$ \vspace{1pt} \paragraph{Proof of 6} \vspace{1pt} Let $A=\left[ a_{ij}\right] _{m\times n}$ and $B=\left[ b_{ij}\right] _{n\times p}$. Then $A^{T}=\left[ a_{ij}^{\prime }\right] _{n\times m}$ and $% B^{T}=\left[ b_{ij}^{\prime }\right] _{p\times n},$ where $a_{ij}^{\prime }=a_{ji}$ and $b_{ij}^{\prime }=b_{ji}.$ The $\left( i,j\right) $ entry of $% B^{T}A^{T}$ is \vspace{1pt} \begin{equation*} \sum_{k=1}^{n}b_{ik}^{\prime }a_{kj}^{\prime }=\sum_{k=1}^{n}b_{ki}a_{jk}=\sum_{k=1}^{n}a_{jk}b_{ki} \end{equation*} This last term is the $\left( j,i\right) $ entry of $AB$ which means that it is the $\left( i,j\right) $ entry of $\left( AB\right) ^{T}.$ \vspace{1pt} \subsection{The Identity Matrix} \vspace{1pt} Consider the $3\times 3$ identity matrix \begin{equation*} I=\left[ \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1% \end{array} \right] \end{equation*} \vspace{1pt} and the matrix \begin{equation*} A=\left[ \begin{array}{cccc} 4 & 1 & 2 & 1 \\ 3 & 0 & 1 & 6 \\ 5 & 7 & 9 & 8% \end{array} \right] \end{equation*} \vspace{1pt} Note the following: \vspace{1pt} \begin{center} \begin{equation*} IA=\allowbreak \left[ \begin{array}{cccc} 4 & 1 & 2 & 1 \\ 3 & 0 & 1 & 6 \\ 5 & 7 & 9 & 8% \end{array}% \right] =A \end{equation*} \vspace{1pt} \end{center} Also,% \begin{equation*} AI=A \end{equation*} \begin{center} \vspace{1pt} \vspace{1pt} \end{center} \subsection{\protect\vspace{1pt}Systems of Equations: Elimination Using Matrices I} \paragraph{\protect\vspace{1pt}Example:} 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 \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 \begin{equation*} AX=C \end{equation*} \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}$\left[ \begin{array}{lllll} 1 & 0 & 0 & 5 & -22 \\ 0 & 1 & 0 & -4 & 17 \\ \mathtt{0} & \mathtt{0} & \mathtt{0} & \mathtt{0} & \mathtt{5}% \end{array} \right] ,$ \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} \begin{equation*} 0x_{1}+0x_{2}+0x_{3}+0x_{4}=5 \end{equation*} \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} \subsection{\protect\vspace{1pt}Elementary Row Operations On Matrices I} \subsubsection{Equivalent Systems} Two linear systems are \textbf{equivalent }if they have the same solutions. \paragraph{Three Elementary Operations} Three basic \emph{elementary }operations are used to transform systems to equivalent systems. These are: \begin{enumerate} \item Interchanging the order of the equations in the system. \item Multiplying any equation by a nonzero constant. \item Replacing any equation in the system by its sum with a nonzero constant multiple of any other equation in the system (elimination step). \end{enumerate} \paragraph{Theorem:} Suppose that an elementary row is performed on a system of linear equations. Then the resulting system has the same set of solutions as the original, so the two systems are equivalent. \vspace{1pt} Operating on the rows of a matrix is equivalent to operating on equations. The row operations that are allowed are the same as the \hyperref{basic operations}{}{}{MA01_02.tex#basic operations} on linear systems of equations: \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}} \end{document} %%%%%%%%%%%%%%%%%%%% End /document/lec_4_24_00.tex %%%%%%%%%%%%%%%%%%%%