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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 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

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

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.

Part 2.

Explain how to describe this solution set using set notation.

Linear Algebra for Team-Based Inquiry Learning 2022 Edition Steven Clontz Drew Lewis University of South Alabama University of South Alabama August 2, 2022