Learning Outcomes
- Determine the number of solutions for a system of linear equations or a vector equation.
Section 1.3 Counting Solutions for Linear Systems (LE3)
Subsection 1.3.1 Class Activities
Activity 1.3.1.
Free browser-based technologies for mathematical computation are available online.
- In the dropdown on the right, you can select a number of different languages. Select "Octave" for the Matlab-compatible syntax used by this text.
- Type
rref([1,3,2;2,5,7])and then press the Evaluate button to compute the \(\RREF\) of \(\left[\begin{array}{ccc} 1 & 3 & 2 \\ 2 & 5 & 7 \end{array}\right]\text{.}\)
Since the vertical bar in an augmented matrix does not affect row operations, the \(\RREF\) of \(\left[\begin{array}{cc|c} 1 & 3 & 2 \\ 2 & 5 & 7 \end{array}\right]\) may be computed in the same way.
Activity 1.3.2.
In the HTML version of this text, code cells are often embedded for your convenience when RREFs need to be computed.
Try this out to compute \(\RREF\left[\begin{array}{cc|c} 2 & 3 & 1 \\ 3 & 0 & 6 \end{array}\right]\text{.}\)
Activity 1.3.3.
Consider the following system of equations.
\begin{alignat*}{4}
3x_1 &\,-\,& 2x_2 &\,+\,& 13x_3 &\,=\,& 6\\
2x_1 &\,-\,& 2x_2 &\,+\,& 10x_3 &\,=\,& 2\\
-x_1 &\,+\,& 3x_2 &\,-\,& 6x_3 &\,=\,& 11\text{.}
\end{alignat*}
(a)
Convert this to an augmented matrix and use technology to compute its reduced row echelon form:
\begin{equation*}
\RREF
\left[\begin{array}{ccc|c}
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\end{array}\right]
=
\left[\begin{array}{ccc|c}
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\end{array}\right]
\end{equation*}
(b)
Use the \(\RREF\) matrix to write a linear system equivalent to the original system.
(c)
How many solutions must this system have?
- Zero
- Only one
- Infinitely-many
Activity 1.3.4.
Consider the vector equation
\begin{equation*}
x_1 \left[\begin{array}{c} 3 \\ 2\\ -1 \end{array}\right]
+x_2 \left[\begin{array}{c}-2 \\ -2 \\ 0 \end{array}\right]
+x_3\left[\begin{array}{c} 13 \\ 10 \\ -3 \end{array}\right]
=\left[\begin{array}{c} 6 \\ 2 \\ 1 \end{array}\right]
\end{equation*}
(a)
Convert this to an augmented matrix and use technology to compute its reduced row echelon form:
\begin{equation*}
\RREF
\left[\begin{array}{ccc|c}
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\end{array}\right]
=
\left[\begin{array}{ccc|c}
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\unknown&\unknown&\unknown&\unknown\\
\end{array}\right]
\end{equation*}
(b)
Use the \(\RREF\) matrix to write a linear system equivalent to the original system.
(c)
How many solutions must this system have?
- Zero
- Only one
- Infinitely-many
Activity 1.3.5.
Is \(0=1\) the only possible logical contradiction obtained from the RREF of an augmented matrix?
- Yes, \(0=1\) is the only possible contradiction from an RREF matrix.
- No, \(0=17\) is another possible contradiction from an RREF matrix.
- No, \(x=0\) is another possible contradiction from an RREF matrix.
- No, \(x=y\) is another possible contradiction from an RREF matrix.
Activity 1.3.6.
Consider the following linear system.
\begin{alignat*}{4}
x_1 &+ 2x_2 &+ 3x_3 &= 1\\
2x_1 &+ 4x_2 &+ 8x_3 &= 0
\end{alignat*}
(a)
Find its corresponding augmented matrix \(A\) and find \(\RREF(A)\text{.}\)
(b)
Use the \(\RREF\) matrix to write a linear system equivalent to the original system.
(c)
How many solutions must this system have?
- Zero
- One
- Infinitely-many
Fact 1.3.7.
We will see in Section 1.4 that the intuition established here generalizes: a consistent system with more nontrivial equations than variables will always have infinitely many solutions.
Fact 1.3.8.
By finding \(\RREF(A)\) from a linear system's corresponding augmented matrix \(A\text{,}\) we can immediately tell how many solutions the system has.
- If the linear system given by \(\RREF(A)\) includes the contradiction \(0=1\text{,}\) that is, the row \(\left[\begin{array}{ccc|c}0&\cdots&0&1\end{array}\right]\text{,}\) then the system is inconsistent, which means it has zero solutions and its solution set is written as \(\emptyset\) or \(\{\}\text{.}\)
- If the linear system given by \(\RREF(A)\) sets each variable of the system to a single value; that is, \(x_1=s_1\text{,}\) \(x_2=s_2\text{,}\) and so on; then the system is consistent with exactly one solution \(\left[\begin{array}{c}s_1\\s_2\\\vdots\end{array}\right]\text{,}\) and its solution set is \(\setList{ \left[\begin{array}{c}s_1\\s_2\\\vdots\end{array}\right] }\text{.}\)
- Otherwise, the system must be consistent with infinitely-many different solutions. We'll learn how to find such solution sets in Section 1.4.
Activity 1.3.9.
For each vector equation, write an explanation for whether each solution set has no solutions, one solution, or infinitely-many solutions. If the set is finite, describe it using set notation.
(a)
\begin{equation*}
x_{1} \left[\begin{array}{c}
1 \\
-1 \\
1
\end{array}\right] + x_{2} \left[\begin{array}{c}
4 \\
-3 \\
1
\end{array}\right] + x_{3} \left[\begin{array}{c}
7 \\
-6 \\
4
\end{array}\right] = \left[\begin{array}{c}
10 \\
-6 \\
4
\end{array}\right]
\end{equation*}
(b)
\begin{equation*}
x_{1} \left[\begin{array}{c}
-2 \\
-1 \\
-2
\end{array}\right] + x_{2} \left[\begin{array}{c}
3 \\
1 \\
1
\end{array}\right] + x_{3} \left[\begin{array}{c}
-2 \\
-2 \\
-5
\end{array}\right] = \left[\begin{array}{c}
1 \\
4 \\
13
\end{array}\right]
\end{equation*}
(c)
\begin{equation*}
x_{1} \left[\begin{array}{c}
-1 \\
-2 \\
1
\end{array}\right] + x_{2} \left[\begin{array}{c}
-5 \\
-5 \\
4
\end{array}\right] + x_{3} \left[\begin{array}{c}
-7 \\
-9 \\
6
\end{array}\right] = \left[\begin{array}{c}
3 \\
1 \\
-2
\end{array}\right]
\end{equation*}
Subsection 1.3.2 Videos
Subsection 1.3.3 Slideshow
Slideshow of activities available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/LE3.slides.html.Exercises 1.3.4 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023e/exercises/#/bank/LE3/.Subsection 1.3.5 Mathematical Writing Explorations
Exploration 1.3.10.
A system of equations with all constants equal to 0 is called homogeneous. These are addressed in detail in section Section 2.9
- Choose three systems of equations from this chapter that you have already solved. Replace the constants with 0 to make the systems homogeneous. Solve the homogeneous systems and make a conjecture about the relationship between the earlier solutions you found and the associated homogeneous systems.
- Prove or disprove. A system of linear equations is homogeneous if an only if it has the the zero vector as a solution.
Subsection 1.3.6 Sample Problem and Solution
Sample problem Example B.1.3.
