## Section1.3Counting Solutions for Linear Systems (LE3)

### Subsection1.3.1Class Activities

#### Activity1.3.1.

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.

#### Activity1.3.2.

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 $$\RREF\left[\begin{array}{cc|c} 2 & 3 & 1 \\ 3 & 0 & 6 \end{array}\right]\text{.}$$

rref([2,3,1;3,0,6])


#### Activity1.3.3.

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?

1. Zero

2. Only one

3. Infinitely-many

rref([3,-2,13,6;2,-2,10,2;-1,3,-6,11])


#### Activity1.3.4.

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?

1. Zero

2. Only one

3. Infinitely-many

rref([3,-2,13,6;2,-2,10,2;-1,0,-3,1])


#### Activity1.3.5.

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.

#### Activity1.3.6.

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?

1. Zero

2. One

3. Infinitely-many

#### Activity1.3.8.

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*}

### Subsection1.3.3Slideshow

Slideshow of activities available at https://teambasedinquirylearning.github.io/linear-algebra/2022/LE3.slides.html.

### Subsection1.3.5Mathematical Writing Explorations

#### Exploration1.3.9.

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.

### Subsection1.3.6Sample Problem and Solution

Sample problem Example B.1.3.