Sharedwww / 54soln1.texOpen in CoCalc
Author: William A. Stein
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%%% First homework for Lenstra's 54.
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%%% AmSLaTeX
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\documentclass[12pt]{article}
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\usepackage{amsmath}
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%\font\bbb=msbm10 scaled \magstep 1
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\newcommand{\prob}[1]{\vspace{.2in}\par\noindent {\bf #1}\par}
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\title{MATH 54\\Linear Algebra and Diff. Eqns.\\
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Section 1 (Lenstra), assignment \# 1}
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\author{William Stein}
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\begin{document}
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\maketitle
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\begin{center}
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{\bf
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{\em Anton}: I.1, \# 5,7,9,13; I.2, \# 3,4ad,5bc,6ad,7bd,12
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}
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\end{center}
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\section{Chapter 1, Section 1}
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\prob{I.1.5}
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(a) \begin{align*}
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2x_1 & = 0\\
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3x_1 - 4x_2 & = 0\\
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x_2 & = 1
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\end{align*}
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(b) \begin{align*}
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3x_1 - 2x_3 & = 5\\
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7x_1 + x_2 +4x_3 & = -3\\
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-2x_2 + x_3 & = 7
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\end{align*}
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(c) \begin{align*}
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7x_1+2x_2+x_3-3x_4 & = 5\\
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x_1+2x_2+4x_3 & = 1
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\end{align*}
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(d) \begin{align*}
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x_1 & = 7\\
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x_2 & = -2\\
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x_3 & = 3\\
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x_4 & = 4
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\end{align*}
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\prob{I.1.7}
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We must find a system of linear equations whose solution $(a,b,c)$
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give the coefficients of $y=ax^2+bx+c$. Since $y=ax^2+bx+c$
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passes through the three indicated points, then by substituting each
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of the three points $(x_1,y_1)$, $(x_2,y_2)$,
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and $(x_3,y_3)$ into the equation $y=ax^2+bx+c$ we see that
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\begin{align*}
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y_1& =ax_1^2+bx_1+c\\
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y_2& =ax_2^2+bx_2+c\\
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y_3& =ax_3^2+bx_3+c.
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\end{align*}
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This system of equations corresponds to the augmented matrix
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\begin{equation*}
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\begin{bmatrix}
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x_1^2&x_1&1&y_1\\
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x_2^2&x_2&1&y_2\\
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x_3^2&x_3&1&y_3
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\end{bmatrix}
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\end{equation*}
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\prob{I.1.9}
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(a) The three lines must not all intersect in a common point.
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For example they could form a "triangle" -- or two could be parallel.
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(b) The three lines all intersect in
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one common point.
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(c) All three of the lines coincide, that is,
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lie on top of each other.
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\prob{I.1.13}
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We must show that if the equations $x_1+kx_2=c$ and
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$x_1+lx_2=d$ have the same solution set, then
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the equations are in fact identical. So assume the
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two equations have the same solution set. To show
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that the equations are the same means to
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show that $k=l$ and $c=d$. To do this we first
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subtract the second equation from the first
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to see that $(k-l)x_2=c-d$. If $k\neq l$
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then $k-l\neq 0$ so we may divide through to
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see that $x_2=(c-d)/(k-l)$. But for every possible value
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of $x_2$ there is a corresponding value of $x_1$
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which provides a solution to $x_1+kx_2=c$ and
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hence to $x_1+lx_2=d$ as well (since they have
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the same solutions). Thus $x_2$ can't only take
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on the value $(c-d)/(k-l)$ so our assumption
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that $k\neq l$ must be false and hence $k=l$.
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But then $c-d=(k-l)x_2=0x_2=0$ so $c=d$ as well.
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This completes the proof.
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\section{Chapter 1, Section 2}
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\prob{I.2.3}
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(a) both
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(b) neither since the second and third rows
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violate property 3
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(c) both
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(d) row-echelon form, but not reduced row-echelon
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form because the second column contains a leading one
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but does not have zeros everywhere else which
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violates property 4
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(e) neither, because property 3 is violated by the
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second row
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(f) both
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\prob{I.2.4ad}
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(a) Once we recover the system of equations from
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its matrix form it is easy to read off the solutions.
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For this matrix, the corresponding system is
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\begin{equation*}
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\begin{align*}
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x_1 + 0 x_2 + 0 x_3& = -3\\
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0x_1 + x_2 + 0 x_3 & = 0 \\
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0x_1 + 0x_2 + x_3 & = 7
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\end{align*}
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\end{equation*}
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so we see at one that the solution is
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$x_1=-3$, $x_2=0$, $x_3=7$.
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(d) The corresponding system of equations is
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\begin{align*}
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x_1-3x_2& = 0\\
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x_3 & = 0\\
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0 & = 1\\
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\end{align*}
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Since the final equation $0=1$ cannot be satisfied
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for any choice if $x_1, x_2, x_3$ we see that this
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system has no solutions.
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\prob{I.2.5bc}
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(b) The corresponding system is
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\begin{align*}
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x_1+8x_3-5x_4&=6\\
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x_2+4x_3-9x_4&=3\\
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x_3+x_4&=2
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\end{align*}
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Solving for the leading variables yields
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\begin{align*}
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x_1 & = 13 x_4 -10 \\
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x_2 & = 13 x_4 - 5 \\
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x_3 & = -x_4 + 2
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\end{align*}
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Since $x_4$ can be assigned an arbitrary value, $t$, there
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are infinitely many solutions. The general solution is
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given by the formulas
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\begin{equation*}
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x_1 = 13t-10\quad x_2=13t-5\quad x_3=-t+2\quad x_4=t.
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\end{equation*}
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(c) The corresponding system is
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\begin{align*}
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x_1+7x_2-2x_3-8x_5&=-3\\
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x_3+x_4+6x_5&=5\\
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x_4+3x_5&=9\\
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0&=0
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\end{align*}
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Solving for the leading variables yields
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\begin{align*}
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x_1&=-65-7x_2-10x_5\\
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x_3&=-4-3x_4\\
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x_4&=9-3x_5
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\end{align*}
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Letting $x_2=s$ and $x_5=t$, the general solution is
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given by the formulas
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\begin{equation*}
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x_1=-11+2t-7s\quad x_2=s\quad x_3 = -4-3t
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\quad x_4 = 9-3t\quad x_5=t
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\end{equation*}
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\prob{I.2.6}
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(a) We first form the augmented matrix
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\begin{equation*}
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\begin{bmatrix}
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1&1&2&8\\
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-1&-2&3&1\\
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3&-7&4&10
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\end{bmatrix}
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\end{equation*}
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Reducing to reduced row-echelon form yields
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\begin{equation*}
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\begin{bmatrix}
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1&0&0&3\\
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0&1&0&1\\
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0&0&1&2\\
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\end{bmatrix}
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\end{equation*}
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(d) The corresponding matrix is
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\begin{equation*}
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\begin{bmatrix}
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0&-2&3&1\\
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3&6&-3&-2\\
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6&6&3&5\\
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\end{bmatrix}
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\end{equation*}
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Reducing to row-echelon form yields
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\begin{equation*}
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\begin{bmatrix}
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1&2&-1&-2/3\\
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0&-2&3&1\\
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0&0&0&6
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\end{bmatrix}
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\end{equation*}
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Since the bottom row corresponds to $0=6$ we
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see the system has no solution.
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\prob{I.2.7}
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(b)
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The corresponding augmented matrix is
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\begin{equation*}
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\begin{bmatrix}
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2&2&2&0\\
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-2&5&2&1\\
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8&1&4&-1
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\end{bmatrix}
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\end{equation*}
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Reducing to row-echelon form yields
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\begin{equation*}
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\begin{bmatrix}
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1&1&1&0\\
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0&7&4&1\\
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0&0&0&0
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\end{bmatrix}
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\end{equation*}
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Back substituting now yields the desired solution
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\begin{equation*}
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x_1=-\frac{3}{7}s-\frac{1}{7}\quad x_2=\frac{1}{7} - \frac{4}{7} s
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\quad x_3 = s
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\end{equation*}
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(d) As in I.2.6, the system is inconsistent.
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\prob{I.2.12}
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(a) Yes, since there are less equations than unknowns.
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(b) No, since there are the same number of equations
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as unknowns and there is no redundancy.
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(c) Yes.
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(b) Yes, since although there are the same number of equations
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as unknowns there is redundancy.
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\end{document}
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