Learning Outcomes
- Compute the solution set for a system of linear equations or a vector equation with infinitely many solutions.
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 1.3.5:
\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 \(\left[\begin{array}{ccc|c}
\markedPivot{1} & 2 & 0 & 4 \\
0 & 0 & \markedPivot{1} & -1
\end{array}\right]\) 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}
\markedPivot{1} & 2 & 0 & 4 \\
0 & 0 & \markedPivot{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):
\begin{equation*}
\unknown = a
\end{equation*}
\begin{equation*}
\unknown = b
\end{equation*}
(c)
Solve for the bound variables (given by the pivot columns) to show that
\begin{equation*}
\unknown = 1+5a+2b
\end{equation*}
\begin{equation*}
\unknown = 1+2a+3b
\end{equation*}
\begin{equation*}
\unknown=3+3b
\end{equation*}
(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}
\hspace{2em}x_1\hspace{2em} \\
\hspace{2em}x_2\hspace{2em} \\
\hspace{2em}x_3\hspace{2em} \\
\hspace{2em}x_4\hspace{2em} \\
\hspace{2em}x_5\hspace{2em}
\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 \(0\) or \(\left[\begin{array}{c}0\\0\\0\end{array}\right]\))
- 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.
Consider the following system of linear equations.
\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*}
(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.
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
Exercises 1.4.3 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2024e/exercises/#/bank/LE4/.Subsection 1.4.4 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 1.4.5 Sample Problem and Solution
Sample problem Example B.1.4.
