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Linear Algebra for Team-Based Inquiry Learning
2022 Edition
| Steven Clontz |
Drew Lewis |
| University of South Alabama |
University of South Alabama |
|
|
August 2, 2022
Section 1.3: Counting Solutions for Linear Systems (LE3)
Activity 1.3.1 (~10 min)
Free browser-based technologies for mathematical computation are available online.
Go to https://sagecell.sagemath.org/.
- 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 (~2 min)
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 (~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 1.3.3 (~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 1.3.3 (~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 1.3.4 (~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 1.3.4 (~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 1.3.4 (~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 1.3.5 (~5 min)
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 (~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 1.3.6 (~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 1.3.6 (~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 1.3.7
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 Linear Systems with Infinitely-Many Solutions (LE4).
Activity 1.3.8 (~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 1.3.8 (~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 1.3.8 (~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*}