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ch2exercises.xml
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ch2exercises.xml
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<?xml version="1.0" encoding="UTF-8"?>
<!--********************************************************************
Copyright 2017 Georgia Institute of Technology
Permission is granted to copy, distribute and/or modify this document
under the terms of the GNU Free Documentation License, Version 1.3 or
any later version published by the Free Software Foundation. A copy of
the license is included in gfdl.xml.
*********************************************************************-->
<section xml:id="ch2exercises">
<title>Exercises</title>
<subsection xml:id="subsection-Rn">
<title>Exercises for Chapter 2</title>
<p>
Questions:
<ol>
<li>
State whether the following are true or false. If false, give a counterexample/reason, and if true, explain why it is true.
<ol>
<li>
A system of 5 linear equations can have 5 solutions.
</li>
<li>
The solution set to any consistent system of linear equations is a subspace.
</li>
<li>
The solution set to a consistent system of linear equations of the form <m>Ax=0</m> is a subspace.
</li>
<li>
The intersection of any two subspaces (i.e., the collection of points that are in both subspaces) is a subspace.
</li>
<li>
The union of any two subspaces (i.e., the collection of points that are in at least on of the subspaces) is a subspace.
</li>
</ol>
</li>
<li>
Use Gaussian elimination to find the unique solution for each system.
<ol>
<li>
<me>\begin{aligned}
2x &+ &3y &= &13 \\
x &- &y &= &-1
\end{aligned}
</me>
</li>
<li>
<me>\begin{aligned}
x & & &- &z &= &0 \\
3x &+ &y & & &= &1 \\
-x &+ &y &+ &z &= &4
\end{aligned}</me>
</li>
</ol>
</li>
<li>
Each system of linear equations is in echelon form (that is, the corresponding matrix is in echelon form).
For each, say whether the system has a unique solution,
no solution, or infinitely many solutions.
<ol>
<li>
<me>
\begin{aligned}
-3x &+ &2y &= &0 \\
& &-2y &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &+ &y & & &= &4 \\
& &y &- &z &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &+ &y & & &= &4 \\
& &y & &-z &= &0 \\
& & & &0 &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &+ &y &= &4 \\
& &0 &= &4
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
3x &+ &6y &+ &z &= &-0.5 \\
& & & &-z &= &2.5 \\
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &- &3y &= &2 \\
& &0 &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
2x &+ &2y &= &4 \\
& &y &= &1 \\
& &0 &= &4
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
2x &+ &y &= &0
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &- &y &= &-1 \\
& &0 &= &0 \\
& &0 &= &4
\end{aligned}
</me>
</li>
<li>
<me>
\begin{aligned}
x &+ &y &- &3z &= &-1 \\
& &y &- &z &= &2 \\
& & & &z &= &0 \\
& & & &0 &= &0
\end{aligned}
</me>
</li>
</ol>
</li>
<li>
Use Gaussian elimination to solve each system
or conclude `many solutions' or `no solutions'.
<ol>
<li><me>
\begin{aligned}
2x &+ &2y &= &5 \\
x &- &4y &= &0
\end{aligned}</me>
</li>
<li><me>
\begin{aligned}
-x &+ &y &= &1 \\
x &+ &y &= &2
\end{aligned}
</me>
</li>
<li><me>
\begin{aligned}
x &- &3y &+ &z &= &1 \\
x &+ &y &+ &2z &= &14
\end{aligned}
</me>
</li>
<li><me>
\begin{aligned}
-x &- &y &= &1 \\
-3x &- &3y &= &2
\end{aligned}
</me>
</li>
<li><me>
\begin{aligned}
& &4y &+ &z &= &20 \\
2x &- &2y &+ &z &= &0 \\
x & & &+ &z &= &5 \\
x &+ &y &- &z &= &10
\end{aligned}
</me>
</li>
<li><me>\begin{aligned}
2x & & &+ &z &+ &w &= &5 \\
& &y & & &- &w &= &-1 \\
3x & & &- &z &- &w &= &0 \\
4x &+ &y &+ &2z &+ &w &= &9
\end{aligned}
</me>
</li>
</ol>
</li>
</ol>
</p>
<p>
Answers:
<ol>
<li>
<ol>
<li>
False. The solution set to any system of linear equations is either empty (inconsistent equations), has a unique solution (consistent equations, no free variables), or infinite (if there is a free variable, since this free variable can then take on the value of any of the infinitely many real numbers).
</li>
<li>
False. For example, the solution set of the set of equations <m>x+y=3</m> in <m>\mathbb R^2</m> is a line, but doesn't pass through the origin.
</li>
<li>
True. Here, we can check the axioms in the definition of a subspace. Given the shape of the equation <m>Ax=0</m>, we clearly have that the zero vector solves this equation, and so the solution set contains the origin (which solves the problem encountered in the last example). Now, suppose that <m>x</m> and <m>y</m> are solutions. Then <m>A(x+y)=Ax+Ay=0+0=0</m>, and so <m>x+y</m> also solves the system, and for a real number <m>\alpha</m>, <m>A(\alpha x)=\alpha Ax=\alpha0=0</m>, and so <m>\alpha x</m> is a solution too. Thus, the solution set satisfies parts 2 and 3 of the definition of a subspace, and hence is a subspace.
</li>
<li>
True. Suppose that <m>V</m> and <m>W</m> are both subspaces of <m>\mathbb R^n</m>. Then we can check each part of the definition is satisfied by the intersection of <m>V</m> and <m>W</m>. Firstly, since 0 is in both subspaces, its in the intersection, so it is a non-empty set. Furthermore, if <m>x</m> and <m>y</m> are in the intersection and <m>\alpha</m> is a real number, then <m>\alpha x+y</m> is in the intersection, using the axioms in the definition of subspace applied twice, once using the fact that <m>x</m> and <m>y</m> are in particular both in <m>V</m>, and once using that they are both in <m>W</m>. Note that this implies both the second and third parts of the definition of subspace, as we can take <m>\alpha=1</m> for part 2 of the definition, and choose <m>y=0</m> for part 3 of the definition.
</li>
<li>
False. Although the union of two intersections always contained the origin, it will usually not be a subspace. For instance, consider the two lines <m>y=\pm x</m> in <m>\mathbb R^2</m>. Their union is not closed under addition. To see this, we just need a counter example. For instance, choose <m>(1,-1)</m> on the line <m>y=-x</m> and the point <m>(2,2)</m> on the line <m>y=x</m>. In particular, both of these points are on the union of these two lines (which looks like an "X" in the plane). However, the sum of these two points, <m>(1,-1)+(2,2)=(3,-1)</m>, is not on the union of the two lines, as <m>3\neq\pm(-1)</m>.
</li>
</ol>
</li>
<li>
<ol>
<li>
Adding <m>-1/2</m> times the first equation to the second equation reduces us to the set of equations
<me>
\begin{aligned}
2x &+ &3y &= &3 \\
&- &(5/2)y &= &-15/2,
\end{aligned}
</me>
giving that the solution is <m>y=3</m> and <m>x=2</m>.
</li>
<li>
Here, we apply Gaussian elimination be adding <m>-3</m> times the first equation to the second equation, and adding the first equation to the second equation to give
<me>
\begin{aligned}
x & & &- &z &= &0 \\
& &y &+ &3z &= &1 \\
& &y & & &= &4
.
\end{aligned}
</me>
We then subtract the second equation from the third:
<me>
\begin{aligned}
x & & &- &z &= &0 \\
& &y &+ &3z &= &1 \\
& & & &-3z &= &3,
\end{aligned}
</me>
giving <m>x=-1</m>, <m>y=4</m>, and <m>z=-1</m>.
</li>
</ol>
</li>
<li>
If a system has a contradictory equation then it has no solution.
Otherwise, if there are any variables that are not leading a row
then it has infinitely many solution.
In the final case, where there is no contradictory equation and
every variable leads some row, it has a unique solution.
<ol>
<li>Unique solution</li>
<li>Infinitely many solutions</li>
<li> Infinitely many solutions</li>
<li> No solution</li>
<li>Infinitely many solutions</li>
<li>Infinitely many solutions</li>
<li>No solution</li>
<li>Infinitely many solutions</li>
<li>No solution</li>
<li>Unique solution</li>
</ol>
</li>
<li>
<ol>
<li>
We add <m>-1/2</m> times the first equation to the second to find
<me>\begin{aligned}
2x &+ &2y &= &5 \\
& &-5y &= &-5/2 ,
\end{aligned}</me>
which shows that <m> y=1/2 </m> and <m> x=2 </m> is the unique solution.
</li>
<li>
Gaussian elimination says that we should add the first and second equations:
<me>\begin{aligned}
-x &+ &y &= &1 \\
& &2y &= &3
\end{aligned}</me>
giving <m> y=3/2 </m> and <m> x=1/2 </m> as the only solution.
</li>
<li>
Row reduction (specifically, subtracting the first equation from the second)
<me>\begin{aligned}
x &- &3y &+ &z &= &1 \\
& &4y &+ &z &= &13
\end{aligned}</me>
shows, because the variable <m>z</m> does not have a pivot in the corresponding column, and hence is a free variable,
that there are many solutions.
</li>
<li>
Row reduction (subtracting 3 times the first equation from the second)
<me>\begin{aligned}
-x &- &y &= &1 \\
& &0 &= &-1
\end{aligned}</me>
shows that there is no solution.
</li>
<li>
Interchanging the first and fourth equations gives
<me>\begin{aligned}
x &+ &y &- &z &= &10 \\
2x &- &2y &+ &z &= &0 \\
x & & &+ &z &= &5 \\
& &4y &+ &z &= &20.
\end{aligned}</me>
Now subtracting the first equation from the third and adding <m>-2</m> times the first equation to the second equation yields
<me>\begin{aligned}
x &+ &y &- &z &= &10 \\
& &-4y &+ &3z &= &-20\\
& &-y &+ &2z &= &-5 \\
& &4y &+ &z &= &20 .
\end{aligned}</me>
Finally, we add the second and fourth equations and subtract <m>1/4</m> times the second row from the third row to obtain
<me>\begin{aligned}
x &+ &y &- &z &= &10 \\
& &-4y &+ &3z &= &-20\\
& & & &(5/4)z &= &0 \\
& & & &4z &= &0,
\end{aligned}</me>
which gives the unique solution<m> (x,y,z)=(5,5,0) </m>.
</li>
<li>
We begin by subtracting twice the first equation from the fourth equation and adding <m>-3/2</m> times the first equation to the third equation to obtain the system
<me>\begin{aligned}
2x & & &+ &z &+ &w &= &5 \\
& &y & & &- &w &= &-1 \\
& & &- &(5/2)z &- &(5/2)w &= &-15/2 \\
& &y & & &- &w &= &-1.
\end{aligned}</me>
Subtracting the second equation from the fourth gives us the system
<me>\begin{aligned}
2x & & &+ &z &+ &w &= &5 \\
& &y & & &- &w &= &-1 \\
& & &- &(5/2)z &- &(5/2)w &= &-15/2 \\
& & & & & &0 &= &0,
\end{aligned}</me>
which shows that there are many solutions.
</li>
</ol>
</li>
</ol>
</p>
</subsection>
</section>