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?

  1. Zero

  2. Only one

  3. 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?

  1. Zero

  2. Only one

  3. Infinitely-many

Activity 1.3.5 (~5 min)

Is \(0=1\) the only possible logical contradiction obtained from the RREF of an augmented matrix?

  1. Yes, \(0=1\) is the only possible contradiction from an RREF matrix.

  2. No, \(0=17\) is another possible contradiction from an RREF matrix.

  3. No, \(x=0\) is another possible contradiction from an RREF matrix.

  4. 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?

  1. Zero

  2. One

  3. 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*}