%%% First homework for Lenstra's 54. %%% AmSLaTeX \documentclass[12pt]{article} \usepackage{amsmath} %\font\bbb=msbm10 scaled \magstep 1 \newcommand{\prob}[1]{\vspace{.2in}\par\noindent {\bf #1}\par} \title{MATH 54\\Linear Algebra and Diff. Eqns.\\ Section 1 (Lenstra), assignment \# 1} \author{William Stein} \begin{document} \maketitle \begin{center} {\bf {\em Anton}: I.1, \# 5,7,9,13; I.2, \# 3,4ad,5bc,6ad,7bd,12 } \end{center} \section{Chapter 1, Section 1} \prob{I.1.5} (a) \begin{align*} 2x_1 & = 0\\ 3x_1 - 4x_2 & = 0\\ x_2 & = 1 \end{align*} (b) \begin{align*} 3x_1 - 2x_3 & = 5\\ 7x_1 + x_2 +4x_3 & = -3\\ -2x_2 + x_3 & = 7 \end{align*} (c) \begin{align*} 7x_1+2x_2+x_3-3x_4 & = 5\\ x_1+2x_2+4x_3 & = 1 \end{align*} (d) \begin{align*} x_1 & = 7\\ x_2 & = -2\\ x_3 & = 3\\ x_4 & = 4 \end{align*} \prob{I.1.7} We must find a system of linear equations whose solution $(a,b,c)$ give the coefficients of $y=ax^2+bx+c$. Since $y=ax^2+bx+c$ passes through the three indicated points, then by substituting each of the three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ into the equation $y=ax^2+bx+c$ we see that \begin{align*} y_1& =ax_1^2+bx_1+c\\ y_2& =ax_2^2+bx_2+c\\ y_3& =ax_3^2+bx_3+c. \end{align*} This system of equations corresponds to the augmented matrix \begin{equation*} \begin{bmatrix} x_1^2&x_1&1&y_1\\ x_2^2&x_2&1&y_2\\ x_3^2&x_3&1&y_3 \end{bmatrix} \end{equation*} \prob{I.1.9} (a) The three lines must not all intersect in a common point. For example they could form a "triangle" -- or two could be parallel. (b) The three lines all intersect in one common point. (c) All three of the lines coincide, that is, lie on top of each other. \prob{I.1.13} We must show that if the equations $x_1+kx_2=c$ and $x_1+lx_2=d$ have the same solution set, then the equations are in fact identical. So assume the two equations have the same solution set. To show that the equations are the same means to show that $k=l$ and $c=d$. To do this we first subtract the second equation from the first to see that $(k-l)x_2=c-d$. If $k\neq l$ then $k-l\neq 0$ so we may divide through to see that $x_2=(c-d)/(k-l)$. But for every possible value of $x_2$ there is a corresponding value of $x_1$ which provides a solution to $x_1+kx_2=c$ and hence to $x_1+lx_2=d$ as well (since they have the same solutions). Thus $x_2$ can't only take on the value $(c-d)/(k-l)$ so our assumption that $k\neq l$ must be false and hence $k=l$. But then $c-d=(k-l)x_2=0x_2=0$ so $c=d$ as well. This completes the proof. \section{Chapter 1, Section 2} \prob{I.2.3} (a) both (b) neither since the second and third rows violate property 3 (c) both (d) row-echelon form, but not reduced row-echelon form because the second column contains a leading one but does not have zeros everywhere else which violates property 4 (e) neither, because property 3 is violated by the second row (f) both \prob{I.2.4ad} (a) Once we recover the system of equations from its matrix form it is easy to read off the solutions. For this matrix, the corresponding system is \begin{equation*} \begin{align*} x_1 + 0 x_2 + 0 x_3& = -3\\ 0x_1 + x_2 + 0 x_3 & = 0 \\ 0x_1 + 0x_2 + x_3 & = 7 \end{align*} \end{equation*} so we see at one that the solution is $x_1=-3$, $x_2=0$, $x_3=7$. (d) The corresponding system of equations is \begin{align*} x_1-3x_2& = 0\\ x_3 & = 0\\ 0 & = 1\\ \end{align*} Since the final equation $0=1$ cannot be satisfied for any choice if $x_1, x_2, x_3$ we see that this system has no solutions. \prob{I.2.5bc} (b) The corresponding system is \begin{align*} x_1+8x_3-5x_4&=6\\ x_2+4x_3-9x_4&=3\\ x_3+x_4&=2 \end{align*} Solving for the leading variables yields \begin{align*} x_1 & = 13 x_4 -10 \\ x_2 & = 13 x_4 - 5 \\ x_3 & = -x_4 + 2 \end{align*} Since $x_4$ can be assigned an arbitrary value, $t$, there are infinitely many solutions. The general solution is given by the formulas \begin{equation*} x_1 = 13t-10\quad x_2=13t-5\quad x_3=-t+2\quad x_4=t. \end{equation*} (c) The corresponding system is \begin{align*} x_1+7x_2-2x_3-8x_5&=-3\\ x_3+x_4+6x_5&=5\\ x_4+3x_5&=9\\ 0&=0 \end{align*} Solving for the leading variables yields \begin{align*} x_1&=-65-7x_2-10x_5\\ x_3&=-4-3x_4\\ x_4&=9-3x_5 \end{align*} Letting $x_2=s$ and $x_5=t$, the general solution is given by the formulas \begin{equation*} x_1=-11+2t-7s\quad x_2=s\quad x_3 = -4-3t \quad x_4 = 9-3t\quad x_5=t \end{equation*} \prob{I.2.6} (a) We first form the augmented matrix \begin{equation*} \begin{bmatrix} 1&1&2&8\\ -1&-2&3&1\\ 3&-7&4&10 \end{bmatrix} \end{equation*} Reducing to reduced row-echelon form yields \begin{equation*} \begin{bmatrix} 1&0&0&3\\ 0&1&0&1\\ 0&0&1&2\\ \end{bmatrix} \end{equation*} (d) The corresponding matrix is \begin{equation*} \begin{bmatrix} 0&-2&3&1\\ 3&6&-3&-2\\ 6&6&3&5\\ \end{bmatrix} \end{equation*} Reducing to row-echelon form yields \begin{equation*} \begin{bmatrix} 1&2&-1&-2/3\\ 0&-2&3&1\\ 0&0&0&6 \end{bmatrix} \end{equation*} Since the bottom row corresponds to $0=6$ we see the system has no solution. \prob{I.2.7} (b) The corresponding augmented matrix is \begin{equation*} \begin{bmatrix} 2&2&2&0\\ -2&5&2&1\\ 8&1&4&-1 \end{bmatrix} \end{equation*} Reducing to row-echelon form yields \begin{equation*} \begin{bmatrix} 1&1&1&0\\ 0&7&4&1\\ 0&0&0&0 \end{bmatrix} \end{equation*} Back substituting now yields the desired solution \begin{equation*} x_1=-\frac{3}{7}s-\frac{1}{7}\quad x_2=\frac{1}{7} - \frac{4}{7} s \quad x_3 = s \end{equation*} (d) As in I.2.6, the system is inconsistent. \prob{I.2.12} (a) Yes, since there are less equations than unknowns. (b) No, since there are the same number of equations as unknowns and there is no redundancy. (c) Yes. (b) Yes, since although there are the same number of equations as unknowns there is redundancy. \end{document}