## Section1.4Linear Systems with Infinitely-Many Solutions (LE4)

### Subsection1.4.1Class Activities

#### Activity1.4.1.

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.

##### (a)

Let $$x_1=a$$ and write the solution set in the form $$\setBuilder { \left[\begin{array}{c} a \\ \unknown \\ \unknown \end{array}\right] }{ a \in \IR } \text{.}$$

##### (b)

Let $$x_2=b$$ and write the solution set in the form $$\setBuilder { \left[\begin{array}{c} \unknown \\ b \\ \unknown \end{array}\right] }{ b \in \IR } \text{.}$$

##### (c)

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?

#### Definition1.4.2.

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.

#### Activity1.4.3.

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.

##### (a)

Row-reduce its augmented matrix.

##### (b)

Assign letters to the free variables (given by the non-pivot columns):

$$\unknown = a$$ and $$\unknown = b\text{.}$$

##### (c)

Solve for the bound variables (given by the pivot columns) to show that

$$\unknown = 1+5a+2b\text{,}$$

$$\unknown = 1+2a+3b\text{,}$$

and $$\unknown=3+3b\text{.}$$

##### (d)

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*}

#### Remark1.4.4.

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 $$\left[\begin{array}{c}1\\2-3a\\a\end{array}\right]$$ ).

#### Activity1.4.5.

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*}

#### Activity1.4.6.

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*}
##### (a)

Explain how to find a simpler system or vector equation that has the same solution set.

##### (b)

Explain how to describe this solution set using set notation.

### Subsection1.4.3Slideshow

Slideshow of activities available at https://teambasedinquirylearning.github.io/linear-algebra/2022/LE4.slides.html.

### Subsection1.4.5Mathematical Writing Explorations

#### Exploration1.4.7.

Construct a system of 3 equations in 3 variables having:

• 0 free variables

• 1 free variable

• 2 free variables

In each case, solve the system you have created. Conjecture a relationship between the number of free variables and the type of solution set that can be obtained from a given system.

#### Exploration1.4.8.

For each of the following, decide if it's true or false. If you think it's true, can we construct a proof? If you think it's false, can we find a counterexample?

• If the coefficient matrix of a system of linear equations has a pivot in the rightmost column, then the system is inconsistent.

• If a system of equations has two equations and four unknowns, then it must be consistent.

• If a system of equations having four equations and three unknowns is consistent, then the solution is unique.

• Suppose that a linear system has four equations and four unknowns and that the coefficient matrix has four pivots. Then the linear system is consistent and has a unique solution.

• Suppose that a linear system has five equations and three unknowns and that the coefficient matrix has a pivot in every column. Then the linear system is consistent and has a unique solution.

### Subsection1.4.6Sample Problem and Solution

Sample problem Example B.1.4.