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Linear Algebra for Team-Based Inquiry Learning
2023 Edition
| Steven Clontz |
Drew Lewis |
| University of South Alabama |
|
|
|
August 24, 2023
Section 1.3: Counting Solutions for Linear Systems (LE3)
Remark 1.3.1
We will frequently need to know the reduced row echelon form of matrices during the remainder of this course, so unless you’re told otherwise, feel free to use technology (see Activity 1.2.13 ) to compute RREFs efficiently.
Activity 1.3.1 (~10 min)
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
\end{alignat*}
Part 1.
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*}
Activity 1.3.1 (~10 min)
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
\end{alignat*}
Part 2.
Use the \(\RREF\) matrix to write a linear system equivalent to the original system.
Activity 1.3.1 (~10 min)
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
\end{alignat*}
Part 3.
How many solutions must this system have?
Zero
Only one
Infinitely-many
Activity 1.3.2 (~10 min)
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*}
Part 1.
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*}
Activity 1.3.2 (~10 min)
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*}
Part 2.
Use the \(\RREF\) matrix to write a linear system equivalent to the original system.
Activity 1.3.2 (~10 min)
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*}
Part 3.
How many solutions must this system have?
Zero
Only one
Infinitely-many
Activity 1.3.3 (~5 min)
What contradictory equations besides \(0=1\) may be obtained from the RREF of an augmented matrix?
\(x=0\) is an obtainable contradiction
\(x=y\) is an obtainable contradiction
\(0=17\) is an obtainable contradiction
\(0=1\) is the only obtainable contradiction
Activity 1.3.4 (~10 min)
Consider the following linear system.
\begin{alignat*}{4}
x_1 &+ 2x_2 &+ 3x_3 &= 1\\
2x_1 &+ 4x_2 &+ 8x_3 &= 0
\end{alignat*}
Part 1.
Find its corresponding augmented matrix \(A\) and find \(\RREF(A)\text{.}\)
Activity 1.3.4 (~10 min)
Consider the following linear system.
\begin{alignat*}{4}
x_1 &+ 2x_2 &+ 3x_3 &= 1\\
2x_1 &+ 4x_2 &+ 8x_3 &= 0
\end{alignat*}
Part 2.
Use the \(\RREF\) matrix to write a linear system equivalent to the original system.
Activity 1.3.4 (~10 min)
Consider the following linear system.
\begin{alignat*}{4}
x_1 &+ 2x_2 &+ 3x_3 &= 1\\
2x_1 &+ 4x_2 &+ 8x_3 &= 0
\end{alignat*}
Part 3.
How many solutions must this system have?
Zero
One
Infinitely-many
Fact 1.3.2
We will see in Section 1.4 Linear Systems with Infinitely-Many Solutions (LE4) that the intuition established here generalizes: a consistent system with more variables than equations (ignoring \(0=0\)) will always have infinitely many solutions.
Fact 1.3.3
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 have more variables than non-trivial equations (equations other than \(0=0\)). This means it is consistent with infinitely-many different solutions. We’ll learn how to find such solution sets in Section 1.4 Linear Systems with Infinitely-Many Solutions (LE4).
Activity 1.3.5 (~15 min)
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.
Part 1.
\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*}
Activity 1.3.5 (~15 min)
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.
Part 2.
\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*}
Activity 1.3.5 (~15 min)
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.
Part 3.
\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*}