\(\newcommand{\circledNumber}[1]{\boxed{#1}}
\newcommand{\IR}{\mathbb{R}}
\newcommand{\IC}{\mathbb{C}}
\renewcommand{\P}{\mathcal{P}}
\renewcommand{\Im}{\operatorname{Im}}
\newcommand{\RREF}{\operatorname{RREF}}
\newcommand{\vspan}{\operatorname{span}}
\newcommand{\setList}[1]{\left\{#1\right\}}
\newcommand{\setBuilder}[2]{\left\{#1\,\middle|\,#2\right\}}
\newcommand{\unknown}{\,{\color{gray}?}\,}
\newcommand{\drawtruss}[2][1]{
\begin{tikzpicture}[scale=#1, every node/.style={scale=#1}]
\draw (0,0) node[left,magenta]{C} --
(1,1.71) node[left,magenta]{A} --
(2,0) node[above,magenta]{D} -- cycle;
\draw (2,0) --
(3,1.71) node[right,magenta]{B} --
(1,1.71) -- cycle;
\draw (3,1.71) -- (4,0) node[right,magenta]{E} -- (2,0) -- cycle;
\draw[blue] (0,0) -- (0.25,-0.425) -- (-0.25,-0.425) -- cycle;
\draw[blue] (4,0) -- (4.25,-0.425) -- (3.75,-0.425) -- cycle;
\draw[thick,red,->] (2,0) -- (2,-0.75);
#2
\end{tikzpicture}
}
\newcommand{\trussNormalForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, blue,->] (4,0) -- (3.5,0.5);
}
\newcommand{\trussCompletion}{
\trussNormalForces
\draw [thick, magenta,<->] (0.4,0.684) -- (0.6,1.026);
\draw [thick, magenta,<->] (3.4,1.026) -- (3.6,0.684);
\draw [thick, magenta,<->] (1.8,1.71) -- (2.2,1.71);
\draw [thick, magenta,->] (1.6,0.684) -- (1.5,0.855);
\draw [thick, magenta,<-] (1.5,0.855) -- (1.4,1.026);
\draw [thick, magenta,->] (2.4,0.684) -- (2.5,0.855);
\draw [thick, magenta,<-] (2.5,0.855) -- (2.6,1.026);
}
\newcommand{\trussCForces}{
\draw [thick, blue,->] (0,0) -- (0.5,0.5);
\draw [thick, magenta,->] (0,0) -- (0.4,0.684);
\draw [thick, magenta,->] (0,0) -- (0.5,0);
}
\newcommand{\trussStrutVariables}{
\node[above] at (2,1.71) {\(x_1\)};
\node[left] at (0.5,0.866) {\(x_2\)};
\node[left] at (1.5,0.866) {\(x_3\)};
\node[right] at (2.5,0.866) {\(x_4\)};
\node[right] at (3.5,0.866) {\(x_5\)};
\node[below] at (1,0) {\(x_6\)};
\node[below] at (3,0) {\(x_7\)};
}
\newcommand{\lt}{<}
\newcommand{\gt}{>}
\newcommand{\amp}{&}
\definecolor{fillinmathshade}{gray}{0.9}
\newcommand{\fillinmath}[1]{\mathchoice{\colorbox{fillinmathshade}{$\displaystyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\textstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptstyle \phantom{\,#1\,}$}}{\colorbox{fillinmathshade}{$\scriptscriptstyle\phantom{\,#1\,}$}}}
\)
Linear Algebra for Team-Based Inquiry Learning
2022 Edition
| Steven Clontz |
Drew Lewis |
| University of South Alabama |
University of South Alabama |
|
|
August 2, 2022
Section 1.4: Linear Systems with Infinitely-Many Solutions (LE4)
Activity 1.4.1 (~10 min)
Consider this simplified linear system found to be equivalent to the system from Activity 1.3.6 :
\begin{alignat*}{3}
x_1 &+ 2x_2 & &= 4\\
& &\phantom{+}x_3 &= -1
\end{alignat*}
Earlier, we determined this system has infinitely-many solutions.
Part 1.
Let \(x_1=a\) and write the solution set in the form \(\setBuilder
{
\left[\begin{array}{c} a \\ \unknown \\ \unknown \end{array}\right]
}{
a \in \IR
}
\text{.}\)
Activity 1.4.1 (~10 min)
Consider this simplified linear system found to be equivalent to the system from Activity 1.3.6 :
\begin{alignat*}{3}
x_1 &+ 2x_2 & &= 4\\
& &\phantom{+}x_3 &= -1
\end{alignat*}
Earlier, we determined this system has infinitely-many solutions.
Part 2.
Let \(x_2=b\) and write the solution set in the form \(\setBuilder
{
\left[\begin{array}{c} \unknown \\ b \\ \unknown \end{array}\right]
}{
b \in \IR
}
\text{.}\)
Activity 1.4.1 (~10 min)
Consider this simplified linear system found to be equivalent to the system from Activity 1.3.6 :
\begin{alignat*}{3}
x_1 &+ 2x_2 & &= 4\\
& &\phantom{+}x_3 &= -1
\end{alignat*}
Earlier, we determined this system has infinitely-many solutions.
Part 3.
Which of these was easier? What features of the RREF matrix \(\left[\begin{array}{ccc|c}
\circledNumber{1} & 2 & 0 & 4 \\
0 & 0 & \circledNumber{1} & -1
\end{array}\right]\) caused this?
Definition 1.4.2
Recall that the pivots of a matrix in \(\RREF\) form are the leading \(1\)s in each non-zero row.
The pivot columns in an augmented matrix correspond to the bound variables in the system of equations (\(x_1,x_3\) below). The remaining variables are called free variables (\(x_2\) below).
\begin{equation*}
\left[\begin{array}{ccc|c}
\circledNumber{1} & 2 & 0 & 4 \\
0 & 0 & \circledNumber{1} & -1
\end{array}\right]
\end{equation*}
To efficiently solve a system in RREF form, assign letters to the free variables, and then solve for the bound variables.
Activity 1.4.3 (~20 min)
Find the solution set for the system
\begin{alignat*}{6}
2x_1&\,-\,&2x_2&\,-\,&6x_3&\,+\,&x_4&\,-\,&x_5&\,=\,&3 \\
-x_1&\,+\,&x_2&\,+\,&3x_3&\,-\,&x_4&\,+\,&2x_5 &\,=\,& -3 \\
x_1&\,-\,&2x_2&\,-\,&x_3&\,+\,&x_4&\,+\,&x_5 &\,=\,& 2
\end{alignat*}
by doing the following.
Part 1.
Row-reduce its augmented matrix.
Activity 1.4.3 (~20 min)
Find the solution set for the system
\begin{alignat*}{6}
2x_1&\,-\,&2x_2&\,-\,&6x_3&\,+\,&x_4&\,-\,&x_5&\,=\,&3 \\
-x_1&\,+\,&x_2&\,+\,&3x_3&\,-\,&x_4&\,+\,&2x_5 &\,=\,& -3 \\
x_1&\,-\,&2x_2&\,-\,&x_3&\,+\,&x_4&\,+\,&x_5 &\,=\,& 2
\end{alignat*}
by doing the following.
Part 2.
Assign letters to the free variables (given by the non-pivot columns):
\(\unknown = a\) and \(\unknown = b\text{.}\)
Activity 1.4.3 (~20 min)
Find the solution set for the system
\begin{alignat*}{6}
2x_1&\,-\,&2x_2&\,-\,&6x_3&\,+\,&x_4&\,-\,&x_5&\,=\,&3 \\
-x_1&\,+\,&x_2&\,+\,&3x_3&\,-\,&x_4&\,+\,&2x_5 &\,=\,& -3 \\
x_1&\,-\,&2x_2&\,-\,&x_3&\,+\,&x_4&\,+\,&x_5 &\,=\,& 2
\end{alignat*}
by doing the following.
Part 3.
Solve for the bound variables (given by the pivot columns) to show that
\(\unknown = 1+5a+2b\text{,}\)
\(\unknown = 1+2a+3b\text{,}\)
and \(\unknown=3+3b\text{.}\)
Activity 1.4.3 (~20 min)
Find the solution set for the system
\begin{alignat*}{6}
2x_1&\,-\,&2x_2&\,-\,&6x_3&\,+\,&x_4&\,-\,&x_5&\,=\,&3 \\
-x_1&\,+\,&x_2&\,+\,&3x_3&\,-\,&x_4&\,+\,&2x_5 &\,=\,& -3 \\
x_1&\,-\,&2x_2&\,-\,&x_3&\,+\,&x_4&\,+\,&x_5 &\,=\,& 2
\end{alignat*}
by doing the following.
Part 4.
Replace \(x_1\) through \(x_5\) with the appropriate expressions of \(a,b\) in the following set-builder notation.
\begin{equation*}
\setBuilder
{
\left[\begin{array}{c}
x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5
\end{array}\right]
}{
a,b\in \IR
}
\end{equation*}
Remark 1.4.4
Don't forget to correctly express the solution set of a linear system. Systems with zero or one solutions may be written by listing their elements, while systems with infinitely-many solutions may be written using set-builder notation.
Inconsistent: \(\emptyset\) or \(\{\}\) (not \(0\)).
Consistent with one solution: e.g. \(\setList{ \left[\begin{array}{c}1\\2\\3\end{array}\right] }\) (not just \(\left[\begin{array}{c}1\\2\\3\end{array}\right]\)).
Consistent with infinitely-many solutions: e.g. \(\setBuilder
{
\left[\begin{array}{c}1\\2-3a\\a\end{array}\right]
}{
a\in\IR
}\) (not just \(\left[\begin{array}{c}1\\2-3a\\a\end{array}\right]\) ).
Activity 1.4.5 (~15 min)
Show how to find the solution set for the vector equation
\begin{equation*}
x_{1} \left[\begin{array}{c}
1 \\
0 \\
1
\end{array}\right] + x_{2} \left[\begin{array}{c}
0 \\
1 \\
-1
\end{array}\right] + x_{3} \left[\begin{array}{c}
-1 \\
5 \\
-5
\end{array}\right] + x_{4} \left[\begin{array}{c}
-3 \\
13 \\
-13
\end{array}\right] = \left[\begin{array}{c}
-3 \\
12 \\
-12
\end{array}\right]\text{.}
\end{equation*}
Activity 1.4.6
Consider the following system of linear equations.
\begin{equation*}
\begin{matrix}
x_{1} & & & - & 2 \, x_{3} & = & -3 \\
5 \, x_{1} & + & x_{2} & - & 7 \, x_{3} & = & -18 \\
5 \, x_{1} & - & x_{2} & - & 13 \, x_{3} & = & -12 \\
x_{1} & + & 3 \, x_{2} & + & 7 \, x_{3} & = & -12 \\
\end{matrix}
\end{equation*}
Part 1.
Explain how to find a simpler system or vector equation that has the same solution set.
Activity 1.4.6
Consider the following system of linear equations.
\begin{equation*}
\begin{matrix}
x_{1} & & & - & 2 \, x_{3} & = & -3 \\
5 \, x_{1} & + & x_{2} & - & 7 \, x_{3} & = & -18 \\
5 \, x_{1} & - & x_{2} & - & 13 \, x_{3} & = & -12 \\
x_{1} & + & 3 \, x_{2} & + & 7 \, x_{3} & = & -12 \\
\end{matrix}
\end{equation*}
Part 2.
Explain how to describe this solution set using set notation.