Learning Outcomes
- Translate back and forth between a system of linear equations, a vector equation, and the corresponding augmented matrix.
Section 1.1 Linear Systems, Vector Equations, and Augmented Matrices (LE1)
Subsection 1.1.1 Class Activities
Definition 1.1.1.
A linear equation is an equation of the variables \(x_i\) of the form
\begin{equation*}
a_1x_1+a_2x_2+\dots+a_nx_n=b\text{.}
\end{equation*}
A solution for a linear equation is a Euclidean vector
\begin{equation*}
\left[\begin{array}{c}
s_1 \\
s_2 \\
\vdots \\
s_n
\end{array}\right]
\end{equation*}
that satisfies
\begin{equation*}
a_1s_1+a_2s_2+\dots+a_ns_n=b
\end{equation*}
(that is, a Euclidean vector that can be plugged into the equation).
Remark 1.1.2.
In previous classes you likely used the variables \(x,y,z\) in equations. However, since this course often deals with equations of four or more variables, we will often write our variables as \(x_i\text{,}\) and assume \(x=x_1,y=x_2,z=x_3,w=x_4\) when convenient.
Definition 1.1.3.
A system of linear equations (or a linear system for short) is a collection of one or more linear equations.
\begin{alignat*}{5}
a_{11}x_1 &\,+\,& a_{12}x_2 &\,+\,& \dots &\,+\,& a_{1n}x_n &\,=\,& b_1 \\
a_{21}x_1 &\,+\,& a_{22}x_2 &\,+\,& \dots &\,+\,& a_{2n}x_n &\,=\,& b_2\\
\vdots& &\vdots& && &\vdots&&\vdots \\
a_{m1}x_1 &\,+\,& a_{m2}x_2 &\,+\,& \dots &\,+\,& a_{mn}x_n &\,=\,& b_m
\end{alignat*}
Its solution set is given by
\begin{equation*}
\setBuilder
{
\left[\begin{array}{c}
s_1 \\
s_2 \\
\vdots \\
s_n
\end{array}\right]
}{
\left[\begin{array}{c}
s_1 \\
s_2 \\
\vdots \\
s_n
\end{array}\right]
\text{is a solution to all equations in the system}
}\text{.}
\end{equation*}
Remark 1.1.4.
When variables in a large linear system are missing, we prefer to write the system in one of the following standard forms:
Original linear system:
\begin{alignat*}{2}
x_1 + 3x_3 &\,=\,& 3\\
3x_1 - 2x_2 + 4x_3 &\,=\,& 0\\
-x_2 + x_3 &\,=\,& -2
\end{alignat*}
Verbose standard form:
\begin{alignat*}{4}
1x_1 &\,+\,& 0x_2 &\,+\,& 3x_3 &\,=\,& 3\\
3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0\\
0x_1 &\,-\,& 1x_2 &\,+\,& 1x_3 &\,=\,& -2
\end{alignat*}
Concise standard form:
\begin{alignat*}{4}
x_1 & & &\,+\,& 3x_3 &\,=\,& 3\\
3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0\\
&\,-\,& x_2 &\,+\,& x_3 &\,=\,& -2
\end{alignat*}
Remark 1.1.5.
It will often be convenient to think of a system of equations as a vector equation.
By applying vector operations and equating components, it is straightforward to see that the vector equation
\begin{equation*}
x_1 \left[\begin{array}{c} 1 \\ 3 \\ 0 \end{array}\right]+ x_2 \left[\begin{array}{c} 0 \\ -2 \\ -1 \end{array}\right]
+ x_3 \left[\begin{array}{c} 3 \\ 4 \\1 \end{array}\right] = \left[\begin{array}{c} 3 \\ 0 \\ -2 \end{array}\right]
\end{equation*}
is equivalent to the system of equations
\begin{alignat*}{4}
x_1 & & &\,+\,& 3x_3 &\,=\,& 3\\
3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0\\
&\,-\,& x_2 &\,+\,& x_3 &\,=\,& -2
\end{alignat*}
Definition 1.1.6.
A linear system is consistent if its solution set is non-empty (that is, there exists a solution for the system). Otherwise it is inconsistent.
Fact 1.1.7.
All linear systems are one of the following:
- Consistent with one solution: its solution set contains a single vector, e.g. \(\setList{\left[\begin{array}{c}1\\2\\3\end{array}\right]}\)
- Consistent with infinitely-many solutions: its solution set contains infinitely many vectors, e.g. \(\setBuilder { \left[\begin{array}{c}1\\2-3a\\a\end{array}\right] }{ a\in\IR }\)
- Inconsistent: its solution set is the empty set, denoted by either \(\{\}\) or \(\emptyset\text{.}\)
Activity 1.1.8.
All inconsistent linear systems contain a logical contradiction. Find a contradiction in this system to show that its solution set is the empty set.
\begin{align*}
-x_1+2x_2 &= 5\\
2x_1-4x_2 &= 6
\end{align*}
Activity 1.1.9.
Consider the following consistent linear system.
\begin{align*}
-x_1+2x_2 &= -3\\
2x_1-4x_2 &= 6
\end{align*}
(a)
Find three different solutions for this system.
(b)
Let \(x_2=a\) where \(a\) is an arbitrary real number, then find an expression for \(x_1\) in terms of \(a\text{.}\) Use this to write the solution set \(\setBuilder
{
\left[\begin{array}{c}
\unknown \\
a
\end{array}\right]
}{
a \in \IR
}\) for the linear system.
Activity 1.1.10.
Consider the following linear system.
\begin{alignat*}{5}
x_1 &\,+\,& 2x_2 &\, \,& &\,-\,& x_4 &\,=\,& 3\\
&\, \,& &\, \,& x_3 &\,+\,& 4x_4 &\,=\,& -2
\end{alignat*}
Describe the solution set
\begin{equation*}
\setBuilder
{
\left[\begin{array}{c}
\unknown \\
a \\
\unknown \\
b
\end{array}\right]
}{
a,b \in \IR
}
\end{equation*}
to the linear system by setting \(x_2=a\) and \(x_4=b\text{,}\) and then solving for \(x_1\) and \(x_3\text{.}\)
Observation 1.1.11.
Solving linear systems of two variables by graphing or substitution is reasonable for two-variable systems, but these simple techniques won’t usually cut it for equations with more than two variables or more than two equations. For example,
\begin{alignat*}{5}
-2x_1 &\,-\,& 4x_2 &\,+\,& x_3 &\,-\,& 4x_4 &\,=\,& -8\\
x_1 &\,+\,& 2x_2 &\,+\,& 2x_3 &\,+\,& 12x_4 &\,=\,& -1\\
x_1 &\,+\,& 2x_2 &\,+\,& x_3 &\,+\,& 8x_4 &\,=\,& 1
\end{alignat*}
has the exact same solution set as the system in the previous activity, but we’ll want to learn new techniques to compute these solutions efficiently.
Remark 1.1.12.
The only important information in a linear system are its coefficients and constants.
Original linear system:
\begin{alignat*}{2}
x_1 + 3x_3 &\,=\,& 3\\
3x_1 - 2x_2 + 4x_3 &\,=\,& 0\\
-x_2 + x_3 &\,=\,& -2
\end{alignat*}
Verbose standard form:
\begin{alignat*}{4}
1x_1 &\,+\,& 0x_2 &\,+\,& 3x_3 &\,=\,& 3\\
3x_1 &\,-\,& 2x_2 &\,+\,& 4x_3 &\,=\,& 0\\
0x_1 &\,-\,& 1x_2 &\,+\,& 1x_3 &\,=\,& -2
\end{alignat*}
Coefficients/constants:
\begin{alignat*}{4}
1 & & 0 &\,\,& 3 &\,|\,& 3\\
3 &\, \,& -2 &\,\,& 4 &\,|\,& 0\\
0 &\, \,& -1 &\,\,& 1 &\,|\,& -2
\end{alignat*}
Definition 1.1.13.
A system of \(m\) linear equations with \(n\) variables is often represented by writing its coefficients and constants in an augmented matrix.
\begin{alignat*}{5}
a_{11}x_1 &\,+\,& a_{12}x_2 &\,+\,& \dots &\,+\,& a_{1n}x_n &\,=\,& b_1\\
a_{21}x_1 &\,+\,& a_{22}x_2 &\,+\,& \dots &\,+\,& a_{2n}x_n &\,=\,& b_2\\
\vdots& &\vdots& && &\vdots&&\vdots\\
a_{m1}x_1 &\,+\,& a_{m2}x_2 &\,+\,& \dots &\,+\,& a_{mn}x_n &\,=\,& b_m
\end{alignat*}
\begin{equation*}
\left[\begin{array}{cccc|c}
a_{11} & a_{12} & \cdots & a_{1n} & b_1\\
a_{21} & a_{22} & \cdots & a_{2n} & b_2\\
\vdots & \vdots & \ddots & \vdots & \vdots\\
a_{m1} & a_{m2} & \cdots & a_{mn} & b_m
\end{array}\right]
\end{equation*}
Example 1.1.14.
The corresponding augmented matrix for this system is obtained by simply writing the coefficients and constants in matrix form.
Linear system:
\begin{alignat*}{2}
x_1 + 3x_3 &\,=\,& 3\\
3x_1 - 2x_2 + 4x_3 &\,=\,& 0\\
-x_2 + x_3 &\,=\,& -2
\end{alignat*}
Augmented matrix:
\begin{equation*}
\left[\begin{array}{ccc|c}
1 & 0 & 3 & 3 \\
3 & -2 & 4 & 0 \\
0 & -1 & 1 & -2
\end{array}\right]
\end{equation*}
Vector equation:
\begin{equation*}
x_1 \left[\begin{array}{c} 1 \\ 3 \\ 0 \end{array}\right]+ x_2 \left[\begin{array}{c} 0 \\ -2 \\ -1 \end{array}\right] + x_3 \left[\begin{array}{c} 3 \\ 4 \\1 \end{array}\right] = \left[\begin{array}{c} 3 \\ 0 \\ -2 \end{array}\right]
\end{equation*}
Subsection 1.1.2 Videos
Subsection 1.1.3 Slideshow
Slideshow of activities available at
https://teambasedinquirylearning.github.io/linear-algebra/2023/LE1.slides.html.Exercises 1.1.4 Exercises
Exercises available at
https://teambasedinquirylearning.github.io/linear-algebra/2023/exercises/#/bank/LE1/.Subsection 1.1.5 Mathematical Writing Explorations
Exploration 1.1.15.
Choose a value for the real constant \(k\) such that the following system has one, many, or no solutions. In each case, write the solution set.
Consider the linear system:
\begin{alignat*}{2}
x_1 - x_2 &\,=\,& 1\\
3x_1 - 3x_2 &\,=\,& k
\end{alignat*}
Exploration 1.1.16.
Consider the linear system:
\begin{alignat*}{2}
ax_1 + bx_2 &\,=\,& j\\
cx_1 + dx_2 &\,=\,& k
\end{alignat*}
- Choose values for \(a,b,c\text{,}\) and \(d\text{,}\) such that \(ad-bc = 0\text{.}\) Show that this system is inconsistent.
- Prove that, if \(ad-bc \neq 0\text{,}\) the system is consistent with exactly one solution.
Exploration 1.1.17.
Given a set \(S\text{,}\) we can define a relation between two arbitrary elements \(a,b \in S\text{.}\) If the two elements are related, we denote this \(a \sim b\text{.}\)
Any relation on a set \(S\) that satisfies the properties below is an equivalence relation.
- Reflexive: For any \(a \in S, a \sim a\)
- Symmetric: For \(a,b \in S\text{,}\) if \(a\sim b\text{,}\) then \(b \sim a\)
- Transitive: for any \(a,b,c \in S, a \sim b \mbox{ and } b \sim c \mbox{ implies } a\sim c\)
For each of the following relations, show that it is or is not an equivalence relation.
- For \(a,b, \in \mathbb{R}\text{,}\) \(a \sim b\) if an only if \(a \leq b\text{.}\)
- For \(a,b, \in \mathbb{R}\text{,}\) \(a \sim b\) if an only if \(|a|=|b|\text{.}\)
Subsection 1.1.6 Sample Problem and Solution
Sample problem Example B.1.1.
