An Inquiry-Based Introduction to Linear Algebra and Applications

Subsection Application: Electrical Circuits

Linear algebra is concerned with the study of systems of linear equations. There are two important aspects to linear systems. One is to use given information to set up a system of equations that represents the information (this is called modeling), and the other is to solve the system. As an example of modeling, we consider the application to the very simple electrical circuit. An electrical circuit consists of

• one or more electrical sources, denoted by

  

• one or more resistors, denoted by

  
   .

A source is a power supply like a battery, and a resistor is an object that consumes the electricity, like a lamp or a computer. A simple circuit consists of one or more sources connected to resistors, like the one shown in Figure 1.1. The straight lines in the circuit indicate wires through which current flows. The points labeled P and Q are called junctions or nodes.

The source creates a charge that produces potential energy \(E\) measured in volts (V). Current flows out of the positive terminal of a source and runs through each branch of the circuit. Let \(I_1\text{,}\) \(I_2\text{,}\) and \(I_3\) be the currents as illustrated in Figure 1.1. The goal is to find the current flowing in each branch of the circuit.

Linear algebra comes into play when analyzing a circuit based on the relationship between current \(I\text{,}\) resistance \(R\text{,}\) and voltage \(E\text{.}\) There are laws governing electrical circuits that state that \(E = IR\) across a resistor. Additionally, Kirchoff's Current and Voltage Laws indicate how current behaves within the whole circuit. Using all these laws together, we derive the system

where \(I_1\text{,}\) \(I_2\text{,}\) and \(I_3\) are the currents at the points indicated in Figure 1.1. To finish analyzing the circuit, we now need to solve this system. In this section we will begin to learn systematic methods for solving systems of linear equations. More details about the derivation of these circuit equations can be found at the end of this section.

Subsection Introduction

Systems of linear equations arise in almost every field of study: mathematics, statistics, physics, chemistry, biology, economics, sociology, computer science, engineering, and many, many others. We will study the theory behind solving systems of linear equations, implications of this theory, and applications of linear algebra as we proceed throughout this text.

Preview Activity 1.1.

(a)

Consider the following system of two linear equations in two unknowns, \(x_1, x_2\text{:}\)

\begin{alignat*}{2} 2x_1 \amp {}-{} \amp 3x_2 \amp = 0\\ x_1 \amp {}-{} \amp x_2 \amp = 1\text{.} \end{alignat*}

One way to solve such a system of linear equations is the method of substitution (where one equation is solved for one variable and then the resulting expression is substituted into the remaining equations). This method works well for simple systems of two equations in two unknowns, but becomes complicated if the number or complexity of the equations is increased.

Another method is elimination — the method that we will adopt in this book. Recall that the elimination method works by multiplying each equation by a suitable constant so that the coefficients of one of the variables in each equation is the same. Then we subtract corresponding sides of these equations to eliminate that variable.

Use the method of elimination to show that this system has the unique solution \(x_1=3\) and \(x_2=2\text{.}\) Explain the specific steps you perform when using elimination.

(b)

Recall that a linear equation in two variables can be represented as a line in \(\R^2\text{,}\) the Cartesian plane, where one variable corresponds to the horizontal axis and the other to the vertical axis. Represent the two equations \(2x_1 -3x_2 = 0\) and \(x_1 - x_2= 1\) in \(\R^2\) and illustrate the solution to the system in your picture.

(c)

The previous example should be familiar to you as a system of two equations in two unknowns. Now we consider a system of three equations in three unknowns

\begin{alignat}{4} {}I_1 \amp {}-{} \amp {}I_2 \amp {}+{} \amp {}I_3 \amp {}={} \amp 0\amp {}\tag{1.1}\\ {5}I_1 \amp {}+{} \amp {2}I_2 \amp {} \amp {} \amp {}={} \amp 8\amp {}\tag{1.2}\\ {} \amp {} \amp {2}I_2 \amp {}+{} \amp {4}I_3 \amp {}={} \amp 5\amp {}\tag{1.3} \end{alignat}

that arises from our electrical circuit in Figure 1.2, with currents \(I_1\text{,}\) \(I_2\text{,}\) and \(I_3\) as indicated in the circuit.

In the remainder of this preview activity we will apply the method of elimination to solve the system of linear equations (1.1), (1.2), and (1.3).

(i)

Replace equation (1.2) with the new equation obtained by multiplying both sides of equation (1.1) by 5 and then subtracting corresponding sides of this equation from the appropriate sides of equation (1.2). Show that the resulting system is

\begin{alignat}{3} {}I_1 \ \amp - \ \amp {}I_2 \ \amp + \ \amp {}I_3 \ \amp = 0\notag\\ \ \amp \ \amp 7I_2 \ \amp {-} \ \amp {5}I_3 \ \amp = 8\tag{1.4}\\ \ \amp {} \ \amp 2I_2 \ \amp + \ \amp 4I_3 \ \amp = 5\text{.}\notag \end{alignat}

(ii)

Now eliminate the variable \(I_2\) from the last two equations in the system in part (a) by using equations (1.3) and (1.4) to show that \(I_3=0.5\text{.}\) Explain your process.

(iii)

Once you know the value for \(I_3\text{,}\) how can you find \(I_2\text{?}\) Then how do you find \(I_1\text{?}\) Use your method to show that the solution to this system is the ordered triple (1,1.5,0.5). Interpret the result in terms of currents.

Subsection Notation and Terminology

To study linear algebra, we will need to agree on some general notation and terminology to represent our systems.

An equation like \(4x_1 + x_2 = 8\) is called a linear equation because the variables (\(x_1\) and \(x_2\) in this case) are raised to the first power, and there are no products of variables. The equation \(4x_1 + x_2 = 8\) is a linear equation in two variables, but we can make a linear equation with any number of variables we like.

We can use any labels for the variables in a linear equation that we like, e.g., \(I_1\text{,}\) \(x_1\text{,}\) \(t_1\text{,}\) and you should become comfortable working with variables in any form. We will usually use subscripts, as in \(x_1, x_2, x_3, \ldots\text{,}\) to represent the variables as this notation allows us to have any number of variables. Other examples of linear equations are

On the other hand, the equations

are non-linear equations.

For example, the two equations

form a system of two linear equations in variables \(x_1\text{,}\) \(x_2\text{.}\)

For example, \(x_1=2, x_2=1\text{,}\) or simply \((2,1)\text{,}\) is a solution to the above system of linear equations in (1.5) as can be checked by substituting the variables into each equation. In solving a system of linear equations, we are interested in finding the set of all solutions, which we will call the solution set of the system. For the above system in (1.5), the solution set is the set containing the single point \((2,1)\text{,}\) denoted \(\{(2,1)\}\text{,}\) because there is only one solution. If we consider just the equation \(x_1-x_2=0\) as our system, the solution set is the line \(x_1=x_2\) in the plane. More generally, a set of solutions is a collection of ordered \(n\)-tuples of numbers. We denote the set of all ordered \(n\)-tuples of numbers as \(\R^n\text{.}\) So, for example, \(\R^2\) is the set of all ordered pairs, or just the standard coordinate plane, and \(\R^3\) is the set of all ordered triples, or the three-dimensional space.

Subsection Solving Systems of Linear Equations

In Preview Activity 1.1, we were introduced to linear systems and the method of elimination for a system of two or three variables. Our goal now is to come up with a systematic method that will reduce any linear system to one that is easy to solve without changing the solution set of the system. Two linear systems will be called equivalent if they have the same solution set.

The operations we used in Preview Activity 1.1 to systematically eliminate variables so that we can solve a linear system are called elementary operations on a system of linear equations or just elementary operations. In the exercises you will argue that elementary operations do not change the solution set to a system of linear equations, a fact that is summarized in the following theorem.

When we apply these elementary operations our ultimate goal is to produce a system of linear equations in a simplified form with the same solution set, where the number of variables eliminated from the equations increase as we move from top to bottom. This method is called the elimination method.

Subsection The Geometry of Solution Sets of Linear Systems

We are familiar with linear equations in two variables from basic algebra and calculus (through linear approximations). The set of solutions to a system of linear equations in two variables has some geometry connected to it.

Activity 1.3 shows that there are three options for the solution set of a system: A system can have no solutions, one solution, or infinitely many solutions.

Now we consider systems of three variables. As an example, let us look at the linear equation \(x+y+z=1\) in the three variables \(x\text{,}\) \(y\text{,}\) and \(z\text{.}\) Notice that the points \((1,1,-2)\text{,}\) \((0,0,0)\text{,}\) and \((-1,-1,2)\) all satisfy this equation. As a linear equation, the graph of \(x+y+z=0\) will be a plane in three dimensions that contains these three points, as shown in Figure 1.7. Hence when we consider a linear system in three unknowns, we are looking for a point in the three dimensional space that lies on all the planes described by the equations.

Subsection Examples

What follows are worked examples that use the concepts from this section.

Subsection Summary

In this section we introduced linear equations and systems of linear equations.

• Informally, a linear equation is an equation in which each term is either a constant or a constant times a variable. More formally, a linear equation in the variables \(x_1\text{,}\) \(x_2\text{,}\) \(\ldots\text{,}\) \(x_n\) is an equation of the form

  \begin{equation*} a_1x_1 + a_2x_2 + \cdots + a_nx_n = b\text{,} \end{equation*}

  where \(n\) is a positive integer and \(a_1\text{,}\) \(a_2\text{,}\) \(\ldots\text{,}\) \(a_n\) and \(b\) are constants.

• A system of linear equations is a collection of one or more linear equations in the same variables.

• Informally, a solution to a system of linear equations is a point that satisfies all of the equations in the system. More formally, a solution to a system of linear equation in \(n\) variables \(x_1\text{,}\) \(x_2\text{,}\) \(\ldots\text{,}\) \(x_n\) is an ordered \(n\)-tuple \((s_1, s_2, \ldots, s_n)\) of numbers so that we obtain all true statements in the system when we replace \(x_1\) with \(s_1\text{,}\) \(x_2\) with \(s_2\text{,}\) \(\ldots\text{,}\) and \(x_n\) with \(s_n\text{.}\)

• Two linear systems are equivalent if they have the same solution set.

• The following operations on a system of equations do not change the solution set:

  1. Replace one equation by the sum of that equation and a scalar multiple of another equation.

  2. Interchange two equations.

  3. Replace an equation by a nonzero scalar multiple of itself.

Subsection Project: Modeling an Electrical Circuit and the Wheatstone Bridge Circuit

Mathematical modeling, or the act of creating equations to model given information, is an important part of problem solving. In this section we will see how we derived the system of equations

to represent the electrical current in the circuit shown in Figure 1.1. Recall that a circuit consists of

• one or more electrical sources (like a battery), denoted by

  

• one or more resistors (like any appliance that you plug into a wall outlet), denoted by

  
  .

The source creates a charge that produces potential energy \(E\) measured in volts (V). No substance conducts electricity perfectly, there is always some price to pay (energy loss) to moving electricity. Electrical current \(I\) in amperes (A) is the flow of the electric charge in the circuit. (A current of 1 ampere means that \(6.2 \times 10^{18}\) electrons pass through the circuit per second.) Current flows out of the positive terminal of a source and runs through each branch of the circuit. Let \(I_1\) be the current flowing through the upper branch, \(I_2\) the current through middle branch, and \(I_3\) the current through the lower branch as illustrated in Figure 1.1. The goal is to find the current flowing in each branch of the circuit.

Linear algebra comes into play when analyzing a circuit based on the relationship between current, resistance, and potential. Three basic principles govern current low in a circuit.

1. Resistance \(R\) in ohms (\(\Omega\)) can be thought of as a measure of how difficult it is to move a charge along a circuit. When a current flows through a resistor, it must expend some energy, called a voltage drop. Ohm's Law states that the voltage drop \(E\) across a resistor is the product of the current \(I\) passing through the resistor and the resistance \(R\text{.}\) That is,

   \begin{equation*} E = IR\text{.} \end{equation*}

2. Kirchoff's Current Law states that at any point in an electrical circuit, the sum of currents flowing into that point is equal to the sum of currents flowing out of that point.

3. Kirchoff's Voltage Law says that around any closed loop the sum of the voltage drops is equal to the sum of the voltage rises.

To see how these laws allow us to model the circuit in Figure 1.1, we will need three equations in \(I_1\text{,}\) \(I_2\text{,}\) and \(I_3\) to determine the values of these currents. Let us first apply Kirchoff's Current Law to the point P. The currents flowing into point P are \(I_1\) and \(I_3\text{,}\) and the current flowing out is \(I_2\text{.}\) This produces the equation \(I_1+I_3 = I_2\text{,}\) or

We have three variables to determine, so we still need two more equations in \(I_1\text{,}\) \(I_2\text{,}\) and \(I_3\text{.}\) Next we apply Kirchoff's Voltage Law to the top loop in the circuit in Figure 1.1. We will assume the following sign conventions:

• A current passing through a resistor produces a voltage drop if it flows in the direction of loop (and a voltage rise if the current passes in the opposite direction of the loop).

• A current passing through a source in the direction of the loop produces a voltage drop if it flows from \(+\) to \(-\) and a voltage rise if it flows from \(-\) to \(+\text{,}\) while a current passing through a source in the opposite direction of the loop produces a voltage rise if it flows from \(+\) to \(-\) and a voltage drop if it flows from \(-\) to \(+\text{.}\)

(The directions chosen in Figure 1.1 for the voltage flow are arbitrary — if we reverse the flow then we just replace voltage drops with voltage rises and obtain the same equations. If a solution shows that a current is negative, then that current flows in the direction opposite of what is shown.)

If we move in the counterclockwise direction around the top loop in the circuit in Figure 1.1, there is a voltage rise through the source of 8 volts. This must equal the voltage drop in this loop. The current \(I_1\) passing though the resistor of resistance 2 \(\Omega\) produces a voltage drop of \(2I_1\) volts. Similarly, the current \(I_1\) passing through the resistor of resistance \(3 \Omega\) produces a voltage drop of \(3I_1\) volts. Finally, the current \(I_2\) passing through the resistor of resistance 2 \(\Omega\) produces a voltage drop of \(2I_2\) volts. So Kirchoff's Voltage Law applied to the top loop in the circuit in Figure 1.1 gives us the equation \(2I_1 + 3I_1 + 2I_2 = 8\) or