**Welcome to the course!** Much of today we’ll discuss
the administ ration of the course .

We will discuss the general description of the course, its
goals and objectives, the syllabus,

the textbook, as signments , quizzes, tests, course grade, and various sundry
items.

** Systems of linear equations.** After discussing the course in general,
we’ll move on the

subject matter. We’re starting out with something you already know, systems of
linear

equations. You’ve looked at these before in various algebra classes, so this is
really a summary .

There are many ways to solve such a system, but we’ll concentrate on a
systematic method

called elimination.

We’ll begin by looking at the the method of elimination as the ancient Chinese
did it.

It’s a standard algorithm that’s been rediscovered over and over. For this
algorithm, a

matrix of numbers is formed from the system of linear equations . For the time
being, the

word “matrix” just means a rectangular array of numbers. As we add operations on
matrices

(“matrices” is the plural of “matrix”), the concept of matrix will mean more
than just a

rectangular array.

Once the matrix of numbers is formed from the system of linear equations, we
will Then

it’s converted into what ’s called “echelon form,” and converted further into
what’s sometimes

called “ reduced echelon form.” From that, the solution can be read directly off
the resulting

matrix.

When there are the same number of equations as there are unknowns , frequently
the

system of equations has exactly one solution . But some times such a system has no
solutions,

and sometimes it has infinitely many solutions.

Matlab. We’ll use Matlab to illustrate topics in linear algebra. It’s available
in the

computer lab.

Our text discusses Matlab in Chapter 12. Read the first
couple of sections for an introduction.

Be low is a short Matlab session to solve the following
system of linear equations.

In the session, a matrix A is created that holds the
coefficients , and a column vector b holds

the constants. The solution is computed with the instruction x = A\b which says
the solution

is (x, y, z) = (1, 2,−2).

**Due Wednesday.** Exercises from section 1.1, page 8,
numbers 1–4, 13–14, 21–22, 23, and

T4.

**Read for Wednesday** sections 1.1 on linear systems and 1.2 on matrices.

Due Friday. Exercises from section 1.2: 1–2, 4–7 parts a–d each, 8, 9, T1,
T5a.