Linear Algebra

72 Chapter 1. Linear Systems
Topic: Analyzing Networks
This is the diagram of an electrical circuit. It happens to describe some of the
connections between a car’s battery and lights, but it is typical of such diagrams.
To read it, we can think of the electricity as coming out of one end of the battery
(labeled 6V OR 12V), flowing through the wires (drawn as straight lines to make
the diagram more readable), and back into the other end of the battery. If, in
making its way from one end of the battery to the other through the network of
wires, some electricity flows through a light bulb (drawn as a circle enclosing a
loop of wire), then that light lights. For instance, when the driver steps on the
brake at point A then the switch makes contact and electricity flows through
the brake lights at point B.
This network of connections and components is complicated enough that to
analyze it — for instance, to find out how much electricity is used when both
the headlights and the brake lights are on — then we need systematic tools.
One such tool is linear systems. To illustrate this application, we first need a
few facts about electricity and networks.
The two facts that we need about electricity concern how the electrical components
act. First, the battery is like a pump for electricity; it provides a force
or push so that the electricity will flow, if there is at least one available path for
it. The second fact about the components is the observation that (in the materials
commonly used in components) the amount of current flow is proportional
to the force pushing it. For each electrical component there is a constant of
proportionality, called its resistance, satisfying that potential = flow·resistance.
(The units are: potential to flow is described in volts, the rate of flow itself is
given in amperes, and resistance to the flow is in ohms. These units are set up
so that volts = amperes · ohms.)
For example, suppose a bulb has a resistance of 25 ohms. Wiring its ends
to a battery with 12 volts results in a flow of electrical current of 12/25 =
0.48 amperes. Conversely, with that same bulb, if we have flow of electrical
current of 2 amperes through it, then the potential difference between one end