# MOOC_LinearAlgebra_Lesson03

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Hello, and welcome to lesson
three of this Introduction to
Linear Algebra with Wolfram U.
The topic for this lesson is
row reduction and echelon forms.
Let’s begin with a brief
overview of the lesson.
So here’s a matrix and
it looks like it's got
pretty random entry, some
are large, some are small.
The goal of this lesson is
to transform such a matrix
to a simpler echelon form.
That's like a much more
ordered set of entries.
So you see over here, this
looks much more orderly
than the earlier matrix.
For example, you got the
ones along over here, and the
last column looks like it's
got some very interesting
and useful information.
The question is, how do
you obtain an echelon form?
Well, an echelon form can
be obtained by using the
process of row reduction.
Fortunately, we have a very
nice function called RowReduce
in the Wolfram Language, which
will give you the reduced
echelon form of a matrix, like
the one you see over here.
And the point is, that once
you have the echelon form
you can solve linear systems.
So, let’s begin with a
formal definition of a
reduced echelon form.
A matrix is said to be in
a reduced echelon form if
it has a few properties.
The first property is that
all non-zero rows are above
any rules with all zeros.
So you can think of the
rows with all zeros as
being kind of redundant.
You do not want them
to be on top at all.
Next, the leading entry in
each non-zero row must be one.
So, you always start with
a one in each non-zero row.
Next, each leading entry
of a row is to the right
of the leading entry
of the row above it.
That’s where the word echelon
comes from, it's like an
orderly arrangement, like
groups of soldiers in an army.
Finally, you want each
leading one to be the only
non-zero entry in its column.
Here's an example of
a matrix that is in a
reduced echelon form.
You can see over here the zeros
are all at the end and then
every rule begins with the one.
If you look at any column
with a one in it, then that's
the only non-zero entry.
The fact is that every
matrix has a unique reduced
echelon form, which can
be found using RowReduce.
Before going into row reduce,
let’s just look at an example
of how you might actually find
the reduced echelon form using
elementary row operation,
just by manipulating the
rows and getting the answer.
Here’s the matrix.
You want to reduce
it to echelon form.
What you do is you take
the second row, R2.
Then you do R2 minus twice R1.
That becomes zero over here.
After that you divide by
-3 to get that to be a 1.
The -3 becomes a 1.
Then he still want to make
that 5 into 0, so you do
rule 1, plus -5, times R2.
So you get a zero over there.
Finally, you do row three.
Goes to R3+R1-R2, and you
get back this nice echelon form.
You can check the answer with
RowReduce and you see that you
get back just the same result.
Let's try and use
row reduce directly.
Here’s a matrix, the matrix has
three rows and four columns.
Apply RowReduce to it.
And sure enough, you get
back a reduced echelon form.
You see over here that this
echelon form is so much
nicer and structured than the
matrix that you began with.
A slightly larger example
now, so here's a random
eight by nine matrix.
I use a SeedRandom just
to make sure I get roughly
the same answer every time.
Here's the random matrix with
eight rows and nine columns.
I apply RowReduce to it
and what you get back is
a very nice, simple form.
Once we know how to use
row reduce, we can now talk
about solving linear systems.
So the question is, how
do you solve a linear
system using row reduction?
Well, the steps are, you first
of all, write on the augmented
matrix with both the left and
right-hand sides of the system.
Apply RowReduce to it and
then, looking at the output
from RowReduce you can
write down the solution set.
In fact, there are
three possibilities.
The first possibility is
that you actually have a
unique solution for your
system, just one solution.
Or you might have infinitely
many solutions and both
of those are set to give
you a consistent system.
The third case is where
you have no solutions and
that's referred to as having
an inconsistent system.
To summarize and make sure we
understand terminology, you can
either have a unique solution
or you can have infinitely
many solutions, or you can
have no solution at all.
One example of each.
Here is a system of equations, a
linear system, three equations,
three unknowns: x, y, and z.
You first of all setup the
augmented matrix system.
Make sure it looks right.
You have 3, 5, 11, 46 in the
first row, that looks correct.
2, -3, 4, 8 second row and 1,
6, -7, -38 in the third row.
All that looks just right.
When you apply RowReduce
to it, look at the matrix
form, and now it's clear that
this first one over here is
just a placeholder for x.
So x is one.
That's a y, so y is two, and
that’s a z, so z is three.
So x=1, y=2, z=3 is a very
nice, simple, unique solution
for this system of equations.
On to a slightly harder
example, we actually have
infinitely many solutions.
Here's a system, right on
the augmented matrix again.
Make sure it looks
right with matrix form.
For example, the first
equation is 5x+2y+11=4.
So you have 5, 2,
11, 4, etcetera.
Apply RowReduce to it.
Look at the matrix form and
you see that the last row
is just a bunch of zeros,
that says you actually have
infinitely many solutions.
The question is, how
do you write them down?
This first row over here
says that x=-25z+10,
because the z goes across
the right-hand side.
y=57z-3.
For each value of z you
actually get a value
of x and a value of y.
For example, if z is one,
then if you plug back over
there, x is -15, and you plug
back over there and y is 34.
You have infinitely
many solutions, each
one corresponding to a
different value of C.
The question is, how did
this example come about?
What I did was, I simply
took the first two equations
over here and I added them, I
summed them to get the 30 here.
So 5+7=12, 2+3=5,
11+4=15, and 4+1=5.
The third equation is
redundant over here, it’s
got no new information and
hence, you actually have
infinitely many solutions to
that system of two equations.
That's one way you might get
infinitely many solutions.
The question is, can we then
go ahead and construct an
example where you have an
inconsistent system like, can
I make this slightly different
to have no solutions at all?
Well I could just make
the five something else
and that's exactly what
I do on the next slide.
Take the first two equations
but I replace the five by a
six, and that's going to give
us an inconsistent system.
The sign of inconsistency
is you have a row that looks
like zeros all the way up
then where you have a one.
That says 0=1, which
is never possible.
Your system, right, the matrix
for it, the augmented matrix,
look at the matrix form and
make sure it's all okay.
It does look okay.
Apply RowReduce, and
you see over here that
the last row is 0=1.
That's no good, there is no
solution for this system,
but of course, we knew that,
we just constructed the way
we wanted to, to make sure
that there is no solutions.
So the six over here made sure
that there is no solution.
So this is a system
which has no solution.
That brings me to the
end of this lesson.
The main point is that every
matrix can be reduced to
an echelon form by using
elementary row operations.
Which are nothing but what
you do if you actually
solved the equation by hand.
You could, let’s say,
interchange two equations or
you might add one multiple of an
equation to another, etcetera.
So those are called
elementary row operations.
Of course, within the Wolfram
Language, we have a nice
function called row reduce,
which will let you do much
more and get to the reduced
echelon form in no time at all.
Now, the point about this
load reduction is, that it can
be used to determine whether
there are any solutions at
all for a system and how
many solutions are there.
So row reduce and row
reduction are a great way
to solve linear systems,
of course, row reduction
is a bit more technical
than using linear solve but
it’s a very powerful way of
understanding linear system.
That's the end of this
lesson and the next lesson
we'll talk about vectors
in n-dimensional space.
Do review this very important
lesson on row reduction and
I hope you've enjoyed it
and learned a lot from it.
Thank you very much,
I stop over here.