Skip to main content

Section 1.4 Linear Systems with Infinitely-Many Solutions (LE4)

Subsection 1.4.1 Class Activities

Activity 1.4.1.

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?

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.

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*}

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.

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*}
(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.

Subsection 1.4.2 Videos

Figure 4. Video: Solving a system of linear equations with infinitely-many solutions

Subsection 1.4.3 Slideshow

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

Exercises 1.4.4 Exercises

Exercises available at https://teambasedinquirylearning.github.io/linear-algebra/2022/exercises/#/bank/LE4/.

Subsection 1.4.5 Mathematical Writing Explorations

Exploration 1.4.7.

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.

Exploration 1.4.8.

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.

Subsection 1.4.6 Sample Problem and Solution

Sample problem Example B.1.4.

You have attempted of activities on this page.