TBIL-LA Linear Systems, Vector Equations, and Augmented Matrices (LE1)

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 $x_{i},$ and assume $x = x_{1}, y = x_{2}, z = x_{3}, w = x_{4}$ when convenient.

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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{aligned} a_{11} x_{1} & + & a_{12} x_{2} & + & \dots & + & a_{1 n} x_{n} & = & b_{1} \\ a_{21} x_{1} & + & a_{22} x_{2} & + & \dots & + & a_{2 n} x_{n} & = & b_{2} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ a_{m 1} x_{1} & + & a_{m 2} x_{2} & + & \dots & + & a_{m n} x_{n} & = & b_{m} \end{aligned}

Its solution set is given by

{[\begin{matrix} s_{1} \\ s_{2} \\ ⋮ \\ s_{n} \end{matrix}] | [\begin{matrix} s_{1} \\ s_{2} \\ ⋮ \\ s_{n} \end{matrix}] is a solution to all equations in the system} .

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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{aligned} x_{1} + 3 x_{3} & = & 3 \\ 3 x_{1} - 2 x_{2} + 4 x_{3} & = & 0 \\ - x_{2} + x_{3} & = & - 2 \end{aligned}

Verbose standard form:

\begin{aligned} 1 x_{1} & + & 0 x_{2} & + & 3 x_{3} & = & 3 \\ 3 x_{1} & - & 2 x_{2} & + & 4 x_{3} & = & 0 \\ 0 x_{1} & - & 1 x_{2} & + & 1 x_{3} & = & - 2 \end{aligned}

Concise standard form:

\begin{aligned} x_{1} & + & 3 x_{3} & = & 3 \\ 3 x_{1} & - & 2 x_{2} & + & 4 x_{3} & = & 0 \\ - & x_{2} & + & x_{3} & = & - 2 \end{aligned}

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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

x_{1} [\begin{matrix} 1 \\ 3 \\ 0 \end{matrix}] + x_{2} [\begin{matrix} 0 \\ - 2 \\ - 1 \end{matrix}] + x_{3} [\begin{matrix} 3 \\ 4 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 0 \\ - 2 \end{matrix}]

is equivalent to the system of equations

\begin{aligned} x_{1} & + & 3 x_{3} & = & 3 \\ 3 x_{1} & - & 2 x_{2} & + & 4 x_{3} & = & 0 \\ - & x_{2} & + & x_{3} & = & - 2 \end{aligned}

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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.

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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. ${[\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]}$
Consistent with infinitely-many solutions: its solution set contains infinitely many vectors, e.g. ${[\begin{matrix} 1 \\ 2 - 3 a \\ a \end{matrix}] | a \in R}$
Inconsistent: its solution set is the empty set, denoted by either ${}$ or $\emptyset .$

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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{aligned} - x_{1} + 2 x_{2} & = 5 \\ 2 x_{1} - 4 x_{2} & = 6 \end{aligned}

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Activity 1.1.9.

Consider the following consistent linear system.

\begin{aligned} - x_{1} + 2 x_{2} & = - 3 \\ 2 x_{1} - 4 x_{2} & = 6 \end{aligned}

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(a)

Find three different solutions for this system.

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(b)

Let $x_{2} = a$ where $a$ is an arbitrary real number, then find an expression for $x_{1}$ in terms of $a .$ Use this to write the solution set ${[\begin{matrix} ? \\ a \end{matrix}] | a \in R}$ for the linear system.

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Activity 1.1.10.

Consider the following linear system.

\begin{aligned} x_{1} & + & 2 x_{2} & - & x_{4} & = & 3 \\ x_{3} & + & 4 x_{4} & = & - 2 \end{aligned}

Describe the solution set

{[\begin{matrix} ? \\ a \\ ? \\ b \end{matrix}] | a, b \in R}

to the linear system by setting $x_{2} = a$ and $x_{4} = b,$ and then solving for $x_{1}$ and $x_{3} .$

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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{aligned} - 2 x_{1} & - & 4 x_{2} & + & x_{3} & - & 4 x_{4} & = & - 8 \\ x_{1} & + & 2 x_{2} & + & 2 x_{3} & + & 12 x_{4} & = & - 1 \\ x_{1} & + & 2 x_{2} & + & x_{3} & + & 8 x_{4} & = & 1 \end{aligned}

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.

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Remark 1.1.12.

The only important information in a linear system are its coefficients and constants.

Original linear system:

\begin{aligned} x_{1} + 3 x_{3} & = & 3 \\ 3 x_{1} - 2 x_{2} + 4 x_{3} & = & 0 \\ - x_{2} + x_{3} & = & - 2 \end{aligned}

Verbose standard form:

\begin{aligned} 1 x_{1} & + & 0 x_{2} & + & 3 x_{3} & = & 3 \\ 3 x_{1} & - & 2 x_{2} & + & 4 x_{3} & = & 0 \\ 0 x_{1} & - & 1 x_{2} & + & 1 x_{3} & = & - 2 \end{aligned}

Coefficients/constants:

\begin{aligned} 1 & 0 & 3 & | & 3 \\ 3 & - 2 & 4 & | & 0 \\ 0 & - 1 & 1 & | & - 2 \end{aligned}

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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{aligned} a_{11} x_{1} & + & a_{12} x_{2} & + & \dots & + & a_{1 n} x_{n} & = & b_{1} \\ a_{21} x_{1} & + & a_{22} x_{2} & + & \dots & + & a_{2 n} x_{n} & = & b_{2} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ a_{m 1} x_{1} & + & a_{m 2} x_{2} & + & \dots & + & a_{m n} x_{n} & = & b_{m} \end{aligned}

[\begin{array}{ccccc} a_{11} & a_{12} & \dots & a_{1 n} & b_{1} \\ a_{21} & a_{22} & \dots & a_{2 n} & b_{2} \\ ⋮ & ⋮ & ⋱ & ⋮ & ⋮ \\ a_{m 1} & a_{m 2} & \dots & a_{m n} & b_{m} \end{array}]

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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{aligned} x_{1} + 3 x_{3} & = & 3 \\ 3 x_{1} - 2 x_{2} + 4 x_{3} & = & 0 \\ - x_{2} + x_{3} & = & - 2 \end{aligned}

Augmented matrix:

[\begin{array}{cccc} 1 & 0 & 3 & 3 \\ 3 & - 2 & 4 & 0 \\ 0 & - 1 & 1 & - 2 \end{array}]

Vector equation:

x_{1} [\begin{matrix} 1 \\ 3 \\ 0 \end{matrix}] + x_{2} [\begin{matrix} 0 \\ - 2 \\ - 1 \end{matrix}] + x_{3} [\begin{matrix} 3 \\ 4 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 0 \\ - 2 \end{matrix}]

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Subsection 1.1.2 Videos

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Figure 1. Video: Converting between systems, vector equations, and augmented matrices

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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{aligned} x_{1} - x_{2} & = & 1 \\ 3 x_{1} - 3 x_{2} & = & k \end{aligned}

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Exploration 1.1.16.

Consider the linear system:

\begin{aligned} a x_{1} + b x_{2} & = & j \\ c x_{1} + d x_{2} & = & k \end{aligned}

Assume

j

and

k

are arbitrary real numbers.

Choose values for $a, b, c,$ and $d,$ such that $a d - b c = 0 .$ Show that this system is inconsistent.
Prove that, if $a d - b c \neq 0,$ the system is consistent with exactly one solution.

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Exploration 1.1.17.

Given a set $S,$ we can define a relation between two arbitrary elements $a, b \in S .$ If the two elements are related, we denote this $a \sim b .$

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 $a, b \in S,$ if $a \sim b,$ then $b \sim a$
Transitive: for any $a, b, c \in S, a \sim b and b \sim c implies a \sim c$

For each of the following relations, show that it is or is not an equivalence relation.

For $a, b, \in R,$ $a \sim b$ if an only if $a \leq b .$
For $a, b, \in R,$ $a \sim b$ if an only if $| a | = | b | .$

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Subsection 1.1.6 Sample Problem and Solution

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Sample problem Example B.1.1.

You have attempted 1 of 1 activities on this page.

Section 1.1 Linear Systems, Vector Equations, and Augmented Matrices (LE1)

Learning Outcomes

Subsection 1.1.1 Class Activities

Definition 1.1.1.

Remark 1.1.2.