\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 4/24/00} \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 $\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| $ $\qquad \qquad \qquad =+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} $ \vspace{1pt} \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:} $\left| \begin{array}{cccc} 10 & -6 & 4 & 9 \\ -2 & 3 & 0 & 6 \\ 1 & -2 & 5 & 10 \\ -6 & 9 & 0 & 3% \end{array} \right| =\allowbreak -1410$ \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 $A=\left[ a_{ij}\right] _{n\times n}=\left[ \begin{array}{llll} a_{11} & a_{12} & . & a_{1n} \\ . & . & . & . \\ . & . & . & . \\ a_{n1} & . & . & a_{nn}% \end{array} \right] _{n\times n}$ \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 \qquad\ \ even\qquad\ \ \ odd \qquad \qquad \paragraph{\protect\vspace{1pt}Example:} $\left| \begin{array}{ll} 1 & -5 \\ 2 & -1% \end{array} \right| =-1+10=9$ \vspace{1pt} \subsection{Minors and Cofactors} \vspace{1pt}We have seen that the formula for a $3\times 3$ determinant is $\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} $ $\vspace{1pt}$We may rewrite the equation above as follows: $\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_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}+a_{12}a_{23}a_{31}-a_{13}a_{22}a_{31}+a_{13}a_{21}a_{32} $ \vspace{1pt} \vspace{1pt}\qquad \qquad \qquad $\ \ \ \ \ \ =a_{11}\left( a_{22}a_{33}-a_{23}a_{32}\right) -a_{12}\left( a_{23}a_{31}-a_{23}a_{31}\right) +a_{13}\left( a_{21}a_{32}-a_{22}a_{31}\right) $ \vspace{1pt} \qquad \qquad \qquad $\ \ \ \ \ =a_{11}\left| \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33}% \end{array} \right| -a_{12}\left| \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33}% \end{array} \right| +a_{13}\left| \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32}% \end{array} \right| .$ \vspace{1pt} Thus \vspace{1pt} $\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( -1\right) ^{1+1}a_{11}\left| \begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33}% \end{array} \right| +\left( -1\right) ^{1+2}a_{12}\left| \begin{array}{ll} a_{21} & a_{23} \\ a_{31} & a_{33}% \end{array} \right| +\left( -1\right) ^{1+3}a_{13}\left| \begin{array}{ll} a_{21} & a_{22} \\ a_{31} & a_{32}% \end{array} \right| $ \paragraph{\protect\vspace{1pt}Example:} $\left| \begin{array}{ccc} 2 & 3 & 5 \\ 1 & 0 & 1 \\ 2 & 1 & 0% \end{array} \right| =\left( -1\right) ^{1+1}\left( 2\right) \left| \begin{array}{cc} 0 & 1 \\ 1 & 0% \end{array} \right| +\left( -1\right) ^{1+2}\left( 3\right) \left| \begin{array}{cc} 1 & 1 \\ 2 & 0% \end{array} \right| +\left( -1\right) ^{1+3}\left( 5\right) \left| \begin{array}{cc} 1 & 0 \\ 2 & 1% \end{array} \right| =2\left( -1\right) -3\left( -2\right) +5\left( 1\right) =9.$ Thus any $3\times 3$ determinant can be written in terms of $2\times 2$ determinants. \vspace{1pt} Definition: The $\left( i,j\right) -$\emph{minor} of an $n\times n$ matrix $% A,$ denoted $M_{ij}\left( A\right) ,$ is defined to be the determinant of the $\left( n-1\right) \times \left( n-1\right) $ matrix formed from $A$ by suppressing all of the elements in the $i$th row and $j$th column of the original matrix $A$. \ The number $C_{ij}\left( A\right) =\left( -1\right) ^{i+j}M_{ij}\left( A\right) $ is called the $\left( i,j\right) -\emph{% cofactor}$ of $A$ and $\left( -1\right) ^{i+j}$ is called the \emph{sign} of the $\left( i,j\right) -$position. \vspace{1pt} \paragraph{Example:} $\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| $ minor of $a_{32}$ is $\left| \begin{array}{ll} a_{11} & a_{13} \\ a_{21} & a_{23}% \end{array} \right| $ \vspace{1pt} Therefore the cofactor $a_{32}$ in the determinant above is $\left( -1\right) ^{3+2}\left| \begin{array}{ll} a_{11} & a_{13} \\ a_{21} & a_{23}% \end{array} \right| $ \vspace{1pt} We have seen that a $3\times 3$ is the sum of the products of the elements of the first row by their corresponding cofactors. Actually one may use any row or column of the determinant to expand it. \vspace{1pt} \paragraph{Example:} Show that the equation \vspace{1pt} \begin{center} \vspace{1pt}\qquad $\left| \begin{array}{lll} x & y & 1 \\ -1 & 2 & 1 \\ 1 & 0 & 1% \end{array} \right| =0$ \end{center} represents a straight line which passes through the points $\left( -1,2\right) $ and $\left( 1,0\right) $. Using cofactors of the second column $\Longrightarrow $ \vspace{1pt} \qquad \qquad \begin{eqnarray*} &&\left( -1\right) ^{1+2}\left( y\right) \left| \begin{array}{ll} -1 & 1 \\ 1 & 1% \end{array} \right| +\left( -1\right) ^{2+2}\left( 2\right) \left| \begin{array}{ll} x & 1 \\ 1 & 1% \end{array} \right| +\left( -1\right) ^{3+2}\left( 0\right) \left| \begin{array}{ll} x & 1 \\ -1 & 1% \end{array} \right| \\ &=&2y+2\left( x-1\right) =0 \end{eqnarray*} Thus the equation of the line is \begin{equation*} x+y=1 \end{equation*} \vspace{1pt} \subsection{Properties of Determinants} \vspace{1pt}We now list the basic properties of determinants. Let $A$ be an $n\times n$ matrix, and let $D_{n}=\left| A\right| $ be its determinant of order $n$. \subsubsection{Theorem 1:} If all elements of any row of $D_{n}$ are zero, then $D_{n}=0$. \vspace{1pt} \paragraph{Example:} Using SNB we see that \begin{center} $\left| \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 \\ -1 & 2 & 10 & 55 \\ -3 & -2 & 5 & 6% \end{array} \right| =\allowbreak 0$ \vspace{1pt} \end{center} which verifies the theorem for this matrix. \vspace{1pt} \subsubsection{Theorem 2:} If $D_{n}$ and $D_{n}^{\prime }$ are two determinants of $n$th order which differ only in that the elements in some row of $D_{n}^{\prime }$ are $k$ times the corresponding elements in $D_{n}$, then \begin{center} \qquad \qquad \begin{equation*} D_{n}^{\prime }=kD_{n} \end{equation*} \vspace{1pt} \end{center} \vspace{1pt} Example. \ $\left| \begin{array}{lll} 1 & 2 & 1 \\ -2 & -8 & 4 \\ 0 & 1 & 5% \end{array} \right| =\allowbreak -26=-2\left| \begin{array}{lll} 1 & 2 & 1 \\ 1 & 4 & -2 \\ 0 & 1 & 5% \end{array} \right| =\allowbreak -2\left( 13\right) =-26$ \vspace{1pt} \subsubsection{Theorem 3:} If $D_{n}^{\prime }$ is obtained from $D_{n}$ by interchanging any pair of rows then \qquad \qquad \qquad \qquad \begin{equation*} D_{n}^{\prime }=-D_{n} \end{equation*} \paragraph{Example:} $\left| \begin{array}{lll} 1 & 2 & 1 \\ 1 & 4 & -2 \\ 0 & 1 & 5% \end{array} \right| =\allowbreak 13$ whereas $\left| \begin{array}{lll} 1 & 4 & -2 \\ 1 & 2 & 1 \\ 0 & 1 & 5% \end{array} \right| =\allowbreak -13$ \subsubsection{Theorem 4:} \vspace{1pt}If two rows of $D_{n}$ are identical, then $D_{n}=0$. \paragraph{Example:} $\left| \begin{array}{ccc} 1 & 2 & 3 \\ 5 & -2 & -1 \\ 1 & 2 & 3% \end{array} \right| =\allowbreak 0$ \vspace{1pt} \subsubsection{Theorem 5:} \vspace{1pt}If two rows of $D_{n}$ are proportional, then $D_{n}=0$. \vspace{1pt} \paragraph{Example:\qquad} $\left| \begin{array}{lll} 1 & 3 & -2 \\ 2 & 4 & 5 \\ -4 & -12 & 8% \end{array} \right| =-4\left| \begin{array}{lll} 1 & 3 & -2 \\ 2 & 4 & 5 \\ 1 & 3 & -2% \end{array} \right| =0$ \subsubsection{Theorem 6:} Suppose each element $a_{kj}$ in the $k$th row of $D_{n}$ is a sum \qquad \qquad \qquad \begin{equation*} a_{kj}=b_{kj}+c_{kj}\qquad j=1,2,...,n \end{equation*} Let $D_{n}^{\prime }$ be obtained from $D_{n}$ by replacing the elements $% a_{kj}$ of the $k$th row by $b_{kj}$ and let $D_{n}^{\prime \prime }$ be obtained from $D_{n}$ by replacing the elements of the $k$th row of $D_{n}$ by $c_{kj}$. Then $D_{n}=D_{n}^{\prime }+D_{n}^{\prime \prime }.$ \vspace{1pt} \paragraph{Example:\ } $\left| \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ b_{21}+c_{21} & b_{22}+c_{22} & b_{23}+c_{23} \\ a_{31} & a_{32} & a_{33}% \end{array} \right| =\left| \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ b_{21} & b_{22} & b_{23} \\ a_{31} & a_{32} & a_{33}% \end{array} \right| +\left| \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ c_{21} & c_{22} & c_{23} \\ a_{31} & a_{32} & a_{33}% \end{array} \right| $ \vspace{1pt} \subsubsection{Theorem 7:} The value of a determinant of order $n$ is unaltered if to each element of any row is added $k$ times the corresponding element of some other row, where $k$ is any constant $\neq 0.$ \vspace{1pt} \paragraph{Example:} $\left| \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31}+ka_{11} & a_{32}+ka_{12} & a_{33}+ka_{13}% \end{array} \right| =\left| \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}% \end{array} \right| +\left| \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ ka_{11} & ka_{12} & ka_{13}% \end{array} \right| =\left| \begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33}% \end{array} \right| +0$ \vspace{1pt} \paragraph{Example:} \vspace{1pt} $\left| \begin{array}{ccc} 1 & 2 & 3 \\ -2 & 0 & 6 \\ 5 & 2 & 1% \end{array} \right| =\allowbreak 40$ Adding $\left( -1\right) $ times row one to row three we have $\left| \begin{array}{ccc} 1 & 2 & 3 \\ -2 & 0 & 6 \\ 0 & -8 & -14% \end{array} \right| =\allowbreak 40$ \vspace{1pt} \subsubsection{Theorem 8:} Let $D_{n}$ be a determinant of order $n$ and let $D_{n}^{\prime }$ be obtained by taking the $1$st, $2$nd, ..., $n$th rows respectively of $D_{n}$ as the $1$st, $2$nd, $n$th columns of $D_{n}^{\prime }$. Then $\vspace{1pt}$ \begin{center} $D_{n}^{\prime }=D_{n}$ \end{center} \paragraph{Remark:} Thus if $A$ is an $n\times n$ matrix, then $\det A=\det A^{t}$ \vspace{1pt} \paragraph{Example:} Evaluate \begin{center} \vspace{1pt}$A=\left| \begin{array}{lll} -1 & 4 & 5 \\ 3 & 6 & 2 \\ 4 & -3 & 0% \end{array} \right| $ \end{center} \vspace{1pt} Let us get another zero in the 3rd row. Multiply second column by $% 4\Longrightarrow $ \vspace{1pt} \begin{center} \qquad $4\det A=\left| \begin{array}{lll} -1 & 16 & 5 \\ 3 & 24 & 2 \\ 4 & -12 & 0% \end{array} \right| $ \end{center} \vspace{1pt} Now add 3 times first column to second. This does not change the value of the determinant above and $\Longrightarrow $ \vspace{1pt} \begin{center} \qquad \qquad \qquad $4\det A=\left| \begin{array}{lll} -1 & 13 & 5 \\ 3 & 33 & 2 \\ 4 & 0 & 0% \end{array} \right| $ \vspace{1pt} \end{center} \vspace{1pt}$4\det A=\left| \begin{array}{lll} -1 & 13 & 5 \\ 3 & 33 & 2 \\ 4 & 0 & 0% \end{array} \right| =\left( 13\right) \left( 2\right) \left( 4\right) -\left( 4\right) \left( 33\right) \left( 5\right) =\allowbreak -556.$ Thus $\det A=-139$ \paragraph{Example:} Evaluate $\det A,$ where $A=\left[ \begin{array}{lll} 3 & -1 & 2 \\ 1 & 6 & -5 \\ -2 & 5 & 4% \end{array} \right] .$ \vspace{1pt} Multiply row 2 by $2$ and add to row 3, row 2 by $-3$ and add to row 1. This does not change the value of the determinant. \vspace{1pt} \vspace{1pt} $\Longrightarrow \det A=\left| \begin{array}{lll} 0 & -19 & 17 \\ 1 & 6 & 5 \\ 0 & 17 & -6% \end{array} \right| =(17)\left( 1\right) \left( 17\right) -(-6)\left( 1\right) \left( -19\right) =\allowbreak 175$ \subsection{Expansion by Cofactors} Definition: The $\left( i,j\right) -$\emph{minor} of an $n\times n$ matrix $% A,$ denoted by $M_{ij}\left( A\right) $ is defined to the the determinant of any order $n$ is the determinant of the $\left( n-1\right) \times \left( n-1\right) $ matrix formed from $A$ by suppressing all the elements in the $% i $th row and $j$th column in the original array. The number $C_{ij}\left( A\right) =\left( -1\right) ^{i+j}M_{ij}\left( A\right) $ is called the $% \left( i,j\right) -\emph{cofactor}$ of $A$ and $\left( -1\right) ^{i+j}$ is called the \emph{sign} of the $\left( i,j\right) -$position. \vspace{1pt} Example. $\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| $ minor of $a_{32}$ is $\left| \begin{array}{ll} a_{11} & a_{13} \\ a_{21} & a_{23}% \end{array} \right| $ \vspace{1pt} Therefore the cofactor $a_{32}$ in the determinant above is $\left( -1\right) ^{3+2}\left| \begin{array}{ll} a_{11} & a_{13} \\ a_{21} & a_{23}% \end{array} \right| $ \subsubsection{Theorem: Laplace Expansion} The determinant of an $n\times n$ matrix $A$ can be computed by using the Laplace expansion along any row or any column of $A.$ More precisely, if $A=% \left[ a_{ij}\right] $ so that $a_{ij}$ is the $\left( i,j\right) -$entry of $A,$ then the expansion along row $i$ is \vspace{1pt} \begin{equation*} \det A=a_{i1}C_{i1}\left( A\right) +a_{i2}C_{i2}\left( A\right) +\cdots +a_{in}C_{in}\left( A\right) \end{equation*} \vspace{1pt} The expansion along column $j$ is given by \vspace{1pt} \begin{equation*} \det A=a_{1j}C_{1j}\left( A\right) +a_{2j}C_{2j}\left( A\right) +\cdots +a_{nj}C_{nj}\left( A\right) \end{equation*} \vspace{1pt} \vspace{1pt} \paragraph{Example:} Evaluate\qquad \qquad \qquad \qquad \qquad $\left| \begin{array}{llll} 1 & 3 & 2 & -1 \\ 2 & -1 & 3 & 1 \\ -1 & 2 & 1 & 1 \\ -2 & -5 & 2 & 3% \end{array} \right| $ \vspace{1pt} $=\left| \begin{array}{llll} 1 & 3 & 2 & -1 \\ 2 & -1 & 3 & 1 \\ -1 & 2 & 1 & 1 \\ -2 & -5 & 2 & 3% \end{array} \right| =\left| \begin{array}{llll} 1 & 3 & 2 & -1 \\ 3 & 2 & 5 & 0 \\ 0 & 5 & 3 & 0 \\ 1 & 4 & 8 & 0% \end{array} \right| =\left( -1\right) ^{1+4}\left( -1\right) \left| \begin{array}{lll} 3 & 2 & 5 \\ 0 & 5 & 3 \\ 1 & 4 & 8% \end{array} \right| =\left| \begin{array}{lll} 0 & -10 & -19 \\ 0 & 5 & 3 \\ 1 & 4 & 8% \end{array} \right| $ $\qquad \qquad \qquad \qquad \qquad \qquad =\left( 1\right) ^{3+1}\left( 1\right) \left| \begin{array}{ll} -10 & -19 \\ 5 & 3% \end{array} \right| =-30+95=65$ \vspace{1pt} \paragraph{Example:} \ $\left| \begin{array}{llll} 2 & 1 & 0 & -1 \\ -5 & 0 & 4 & 2 \\ 1 & -3 & 0 & 4 \\ 0 & 0 & -1 & -2% \end{array} \right| =\left( -1\right) ^{4+3}\left( -1\right) \left| \begin{array}{lll} 2 & 1 & -1 \\ -5 & 0 & 2 \\ 1 & -3 & 4% \end{array} \right| +\left( -2\right) \left( -1\right) ^{4+4}\left| \begin{array}{lll} 2 & 1 & 0 \\ -5 & 0 & 4 \\ 1 & -3 & 0% \end{array} \right| $ \vspace{1pt} \subsection{More about Determinants} \subsubsection{\protect\vspace{1pt}Theorem:} If $A$ is an $n\times n$ matrix, then $\det \left( kA\right) =k^{n}\det A$ for any real number $k.$ \vspace{1pt} \paragraph{Example:} $\left| \begin{array}{ccc} 2 & 4 & 6 \\ 10 & 12 & 14 \\ 6 & 10 & 12% \end{array} \right| =\allowbreak 32=2^{3}\left| \begin{array}{ccc} 1 & 2 & 3 \\ 5 & 6 & 7 \\ 3 & 5 & 6% \end{array} \right| =2^{3}(4)$ Definition: If $A=\left[ a_{ij}\right] _{n\times n},$ then the elements $% a_{ii}$ form the \emph{main diagonal} of $A.$ $A$ is said to be \emph{lower (upper) triangular} if all the entries above (below) the main diagonal are zero. \subsubsection{Theorem:} If $A$ is a square triangular matrix, the $\det A$ is the product of the entries on the main diagonal. \vspace{1pt} \paragraph{Example:} $\left| \begin{array}{cccc} -1 & 0 & 0 & 0 \\ -4 & 2 & 0 & 0 \\ 10 & 15 & -3 & 0 \\ 12 & 34 & 32 & 1% \end{array} \right| =\allowbreak 6=\left( -1\right) \left( 2\right) \left( -3\right) \left( 1\right) $ \vspace{1pt} \subsubsection{Theorem:} If $A$ and $B$ are $n\times n$ matrices, then \vspace{1pt} \begin{equation*} \det \left( AB\right) =\det A\det B \end{equation*} \vspace{1pt} \paragraph{Example:} $\det \left[ \begin{array}{cc} 1 & 3 \\ -1 & 5% \end{array} \right] =\allowbreak 8\qquad \det \left[ \begin{array}{cc} 2 & -3 \\ -1 & 4% \end{array} \right] =\allowbreak 40$ \vspace{1pt} $\left[ \begin{array}{cc} 1 & 3 \\ -1 & 5% \end{array} \right] \left[ \begin{array}{cc} 2 & -3 \\ -1 & 4% \end{array} \right] =\allowbreak \left[ \begin{array}{cc} -1 & 9 \\ -7 & 23% \end{array} \right] $ \ and $\det \left[ \begin{array}{cc} -1 & 9 \\ -7 & 23% \end{array} \right] =\allowbreak 40$ \vspace{1pt} \subsubsection{Theorem:} If $A$ is a square matrix, then \begin{equation*} \det \left( A^{k}\right) =\left( \det A\right) ^{k} \end{equation*} \vspace{1pt} \subsection{Cramer's Rule} \vspace{1pt}We have seen the system of $n$ equations in $n$ unknowns \vspace{1pt} \begin{eqnarray*} a_{11}x_{1}+a_{12}x_{2}+\cdots +a_{nn}x_{n} &=&b_{1} \\ a_{21}x_{1}+a_{22}x_{2}+\cdots +a_{2n}x_{n} &=&b_{2} \\ \cdots &=&\cdots \\ a_{n1}x_{1}+a_{n2}x_{2}+\cdots +a_{nn}x_{n} &=&b_{n} \end{eqnarray*} may be rewritten in the form \vspace{1pt} \begin{equation*} AX=B \end{equation*} \vspace{1pt} where $A=\left[ a_{ij}\right] _{n\times n},$ $X=\left[ \begin{array}{c} x_{1} \\ x_{2} \\ \vdots \\ x_{n}% \end{array} \right] $ and $B=\left[ \begin{array}{c} b_{1} \\ b_{2} \\ \vdots \\ b_{n}% \end{array} \right] .$ \subsubsection{Theorem (Cramer's Rule)} \vspace{1pt}If $A$ is an invertible $n\times n$ matrix, the solution to the system \vspace{1pt} \begin{equation*} AX=B \end{equation*} \vspace{1pt} of $n$ equations in $n$ unknowns $x_{1},x_{2},....,x_{n}$ is given by \vspace{1pt} \begin{center} $x_{1}=\dfrac{\det A_{1}}{\det A},\qquad x_{2}=\dfrac{\det A_{2}}{\det A}% ,\qquad \ldots ,\qquad x_{n}=\dfrac{\det A_{n}}{\det A}$ \vspace{1pt} \end{center} where, for each $k$, $A_{k}$ is the matrix obtained from $A$ by replacing the $kth$ column by the entries in $B.$ \vspace{1pt} \paragraph{Example:} Use Cramer's Rule to find $x_{1}$. \vspace{1pt} \begin{eqnarray*} 5x_{1}+x_{2}-x_{3} &=&4 \\ 9x_{1}+x_{2}-x_{3} &=&1 \\ x_{1}-x_{2}+5x_{3} &=&2 \end{eqnarray*} \vspace{1pt} $A=\left[ \begin{array}{ccc} 5 & 1 & -1 \\ 9 & 1 & -1 \\ 1 & -1 & 5% \end{array} \right] $ and $A_{1}=\left[ \begin{array}{ccc} 4 & 1 & -1 \\ 1 & 1 & -1 \\ 2 & -1 & 5% \end{array} \right] .$ Then $\det A=\det \left[ \begin{array}{ccc} 5 & 1 & -1 \\ 9 & 1 & -1 \\ 1 & -1 & 5% \end{array} \right] =\allowbreak -16$ and $\det A_{1}=\det \left[ \begin{array}{ccc} 4 & 1 & -1 \\ 1 & 1 & -1 \\ 2 & -1 & 5% \end{array} \right] =\allowbreak 12$ so $x_{1}=\dfrac{12}{-16}=\allowbreak -\frac{3}{4}.$ \vspace{1pt} Similarly $\det A_{2}=\left| \begin{array}{ccc} 5 & 4 & -1 \\ 9 & 1 & -1 \\ 1 & 2 & 5% \end{array} \right| =\allowbreak -166$ so $x_{2}=\dfrac{-166}{-16}=\allowbreak \frac{83}{% 8}.$ One may check these values by using SNB Maple, Solve, Exact. \end{document} %%%%%%%%%%%%%%%%%%%% End /document/lec_4_24_00.tex %%%%%%%%%%%%%%%%%%%%