\hfill \thepage} %} %\newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} %\newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} %\newtheorem{conjecture}[theorem]{Conjecture} %\newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} %\newtheorem{definition}[theorem]{Definition} %\newtheorem{example}[theorem]{Example} %\newtheorem{exercise}[theorem]{Exercise} %\newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} %\newtheorem{proposition}[theorem]{Proposition} %\newtheorem{remark}[theorem]{Remark} \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 5/01/00} \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} \paragraph{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 \begin{equation*} 4\det A=\left| \begin{array}{lll} -1 & 13 & 5 \\ 3 & 33 & 2 \\ 4 & 0 & 0% \end{array}% \right| \end{equation*} \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 \begin{equation*} A=\left[ \begin{array}{lll} 3 & -1 & 2 \\ 1 & 6 & -5 \\ -2 & 5 & 4% \end{array}% \right] \end{equation*}% $.$ \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:} \begin{equation*} \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) \end{equation*} 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:} \begin{equation*} \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) \end{equation*} \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{More Examples} \paragraph{\protect\vspace{1pt}Example:} Let \vspace{1pt} \qquad \qquad \qquad \begin{equation*} B=\left[ \begin{array}{ll} 2 & 5 \\ 1 & 3% \end{array}% \right] \end{equation*} \ Find $B^{-1}$.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \vspace{1pt} \paragraph{\protect\vspace{1pt}SOLUTION:} Suppose $B^{-1}=\left[ \begin{array}{ll} a_{1} & a_{2} \\ b_{1} & b_{2}% \end{array} \right] $ . \vspace{1pt} \vspace{1pt}Then we want \vspace{1pt} \vspace{1pt}\qquad \qquad \qquad \qquad $\left[ \begin{array}{ll} 2 & 5 \\ 1 & 3% \end{array} \right] \left[ \begin{array}{ll} a_{1} & a_{2} \\ b_{1} & b_{2}% \end{array} \right] =\left[ \begin{array}{ll} 1 & 0 \\ 0 & 1% \end{array} \right] $ \vspace{1pt}$\Longrightarrow 2a_{1}+5b_{1}=1,\qquad 2a_{2}+5b_{2}=0\qquad 1\ast a_{1}+3b_{1}=0\qquad 1\ast a_{2}+3b_{2}=1$ Hence $b_{1}=-1,\qquad b_{2}=2,\qquad a_{1}=3,$ and $a_{2}=-5$. \vspace{1pt} $B^{-1}=\left[ \begin{array}{ll} a_{1} & a_{2} \\ b_{1} & b_{2}% \end{array} \right] =\left[ \begin{array}{cc} 3 & -5 \\ -1 & 2% \end{array} \right] $ \paragraph{Example:} \vspace{1pt} For what values, if any, of $k_{1}$ and $k_{2}$ does the system \qquad $\qquad $% \begin{eqnarray*} 2x_{1}+4x_{2}-x_{3}+x_{4} &=&0 \\ 4x_{1}-2x_{2}+x_{3}+x_{4} &=&k_{1} \\ 2x_{1}+14x_{2}-4x_{3}+2x_{4} &=&k_{2} \end{eqnarray*} \vspace{1pt}$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \qquad $ \vspace{1pt}have a solution? Justify your conclusion. You need not solve the system if it has a solution. \paragraph{\protect\vspace{1pt}SOLUTION} The augmented matrix is: \begin{equation*} \left[ \begin{array}{llll} 2 & -4 & -1 & 1 \\ 4 & -2 & 1 & 1 \\ 2 & 14 & -4 & 2% \end{array}% \begin{array}{l} 0 \\ k_{1} \\ k_{2}% \end{array}% \right] \end{equation*} Subtracting row $1$ from row $3$ and $2\times $ row $1$ from row $2$ we obtain \begin{equation*} \left[ \begin{array}{llll} 2 & -4 & -1 & 1 \\ 0 & -10 & 3 & -1 \\ 0 & 10 & -3 & 1% \end{array}% \begin{array}{l} 0 \\ k_{1} \\ k_{2}% \end{array}% \right] \end{equation*} Now add row $2$ to row $3$: \begin{equation*} \left[ \begin{array}{llll} 2 & -4 & -1 & 1 \\ 0 & -10 & 3 & -1 \\ 0 & 0 & 0 & 0% \end{array}% \begin{array}{l} 0 \\ k_{1} \\ k_{2}+k_{1}% \end{array}% \right] \end{equation*} We see that that there is no solution unless .% \begin{equation*} k_{2}+k_{1}=0, \end{equation*} or $k_{2}=-k_{1}.$ The given system has a solution if and only if the above condition is fulfilled In our case, there are infinitely many solutions when $% k_{2}=-k_{1}$. \paragraph{} \paragraph{Example:} Find det $A$ if \vspace{1pt} \qquad \begin{equation*} A=\left[ \begin{array}{llll} 4 & 0 & -2 & 0 \\ 1 & 1 & 2 & 0 \\ 1 & 3 & 5 & 6 \\ 1 & 0 & 0 & -3% \end{array}% \right] \end{equation*}% \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \paragraph{\protect\vspace{1pt}SOLUTION} \begin{equation*} A=\left[ \begin{array}{llll} 4 & 0 & -2 & 0 \\ 1 & 1 & 2 & 0 \\ 1 & 3 & 5 & 6 \\ 1 & 0 & 0 & -3% \end{array}% \right] \end{equation*} \ Subtract row $2$ from row $3$: \begin{equation*} A=\left[ \begin{array}{llll} 4 & 0 & -2 & 0 \\ 1 & 1 & 2 & 0 \\ 0 & 2 & 3 & 6 \\ 1 & 0 & 0 & -3% \end{array}% \right] \end{equation*} \ Multiply column $2$ by $2$ and subtract from column $3$: \begin{equation*} A=\left[ \begin{array}{llll} 4 & 0 & -2 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 2 & -1 & 6 \\ 1 & 0 & 0 & -3% \end{array}% \right] \end{equation*} \ Add column $3$ to column $2$: \begin{equation*} A=\left[ \begin{array}{llll} 4 & -2 & -2 & 0 \\ 1 & 1 & 0 & 0 \\ 0 & 1 & -1 & 6 \\ 1 & 0 & 0 & -3% \end{array}% \right] \end{equation*} \ Subtract row $2$ from row $3$ and add it twice to row $1$:% \begin{equation*} A=\left[ \begin{array}{llll} 6 & 0 & -2 & 0 \\ 1 & 1 & 0 & 0 \\ -1 & 0 & -1 & 6 \\ 1 & 0 & 0 & -3% \end{array}% \right] \end{equation*} \ Expand by second column: \vspace{1pt} \begin{equation*} A=+1\times \left[ \begin{array}{lll} 6 & -2 & 0 \\ -1 & -1 & 6 \\ 1 & 0 & -3% \end{array}% \right] \end{equation*} Now $\det A=18+6-12=12.$ \vspace{1pt} \paragraph{\protect\vspace{1pt}Example:} Evaluate \vspace{1pt} \begin{equation*} \det \left| \begin{array}{lllll} 1 & 2 & 3 & 4 & 5 \\ 6 & 7 & 8 & 9 & 10 \\ 11 & 12 & 13 & 14 & 15 \\ 16 & 17 & 18 & 19 & 20 \\ 21 & 22 & 23 & 24 & 25% \end{array}% \right| \end{equation*}% $\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad $ \vspace{1pt} \paragraph{SOLUTION} \vspace{1pt} Here is one of the many ways to do this. Subtract column $1$ from columns $2$ and $5$: $\vspace{1pt}$ $\left[ \begin{array}{lllll} 1 & 1 & 3 & 4 & 4 \\ 6 & 1 & 8 & 9 & 4 \\ 11 & 1 & 13 & 14 & 4 \\ 16 & 1 & 18 & 19 & 4 \\ 21 & 1 & 23 & 24 & 4% \end{array} \right] $ \vspace{1pt} Column $2$ multiplied by $4$ gives column $5$. Consequently, the two columns are proportional, so the value of the determinant is $0$. \end{document} %%%%%%%%%%%%%%%%%%%%% End /document/lec_5_1_00.tex %%%%%%%%%%%%%%%%%%%%