BEN HARRIS: Hi.
I'm Ben.
Today we are going to do an
LU decomposition problem.
Here's the problem right here.
Find that LU decomposition
of this matrix A.
Now notice that this matrix
A has variables, as well
as numbers.
So the sentence
ends: when it exists.
And the second part
of the question
asks you: for which
real numbers a and b
does the LU decomposition of
this matrix actually exist?
Now, you can hit pause now and
I'll give you a few seconds.
You can try to solve
this on your own,
and then we'll be back
and we can do it together.
And we're back.
Now, what do you have to
remember when doing an LU
decomposition problem?
Well, we do elimination
in the same way
that we did before in order to
find U. But with this question
we need to find L as well.
So we need to do
elimination, but we also
need to keep track of
the elimination matrices
along the way.
Good.
So let's do that.
So let me put my matrix up here.
And we want to do elimination.
So which entry do
we eliminate first?
That's right.
It's this (2, 1) entry.
So we replace the second
row by the second row
minus a times the first
row, and we get this.
But we're not just
doing elimination,
we're finding an
LU decomposition.
So we need to keep
track of the matrix
that I multiplied the
elimination matrix, that I
multiplied this matrix A by on
the left to get this matrix.
So what is that?
That's this E_(2,1).
Since I eliminated the (2, 1)
entry, I'll call it E_(2,1).
And it's this matrix.
Why is it this matrix?
Well, remember how
multiplication on the left
works.
I replaced the first row
by just the first row.
I replaced the second
row by the second row
minus a times the first row.
So you can just read
off from these rows
which operations I did.
Now, which entries
should we eliminate next?
We need to eliminate this b.
So we will replace the
third row by the third row
minus b times the first row.
And which elimination
matrix did we use?
Well, note, we
replaced the third row
by the third row minus
b times the first row.
That's exactly what you should
read off this elimination
matrix.
Good.
Now, we only have one step left.
We only need to
eliminate one last entry.
But this one's a little
tricky, so let's be careful.
In order to eliminate this
b, we need a to be a pivot.
In particular, we
need a to be nonzero.
If a were zero
here, then we would
have to do a row exchange.
And that's no good.
You can't find an
LU decomposition
if you have to do a row
exchange in elimination.
So we need to assume
that a is non-zero
in order to keep going.
So let's just assume
there that a is non-zero.
Now, what do we do?
Well we can replace
the third row
by the third row minus b
over a times the second row.
And we just get this.
a minus b.
And let's write down
our elimination matrix.
E_(3,2) now.
There's our elimination matrix.
We replaced the third row
by the third row minus b
over a times the second row.
Good.
So we found our U matrix.
That's what elimination does,
it gives us our U matrix.
So let me write it up here.
1, 0, 1; 0, a, 0;
0, 0, a minus b.
Good.
Now we have to
find our L matrix,
and we need to use these
elimination matrices
that we've been recording along
the way in order to do that.
So remember that we started
with A, and we got U.
And how did we do that?
Well we multiplied on the left
by all of these elimination
matrices, E_(2,1),
E_(3,1) and E_(3,2).
Sorry if that's
scrunching together there.
Now, if we move these
elimination matrices
to the other side then we'll
get L. So what do we have?
We have A equals E_(2,1)
inverse, E_(3,1) inverse,
E_(3,2) inverse times
U. And this is our L,
this product of
these three matrices.
Good.
So let's compute it now.
So L is the product
of three matrices.
I need to get them by going
back and looking at these three
elimination matrices and
taking their inverses.
Well the nice thing
about taking an inverse
of an elementary
matrix like this is we
just make a minus a
plus or a plus a minus.
So that's easy enough.
We just change the
off-diagonal entries,
we just change their signs.
You can check that that
does what we wanted it to.
It gives us the inverse.
Good.
And the last comment is
that multiplying these three
matrices is really
easy in this order.
Turns out all you do is you just
plop these entries right in.
1, 1.
Good.
So this is our L matrix.
So now we have our U
matrix and our L matrix,
and we're done with the
first part of the question.
The second part asks us for
which real numbers a and b
does this decomposition exist?
Now let's go back and
remember that at one point
we had to assume
that A was non-zero.
That was the only
assumption we had to make
to get this decomposition.
So it exists-- it being
this decomposition--
when a is non-zero.
And that's the answer
to the second part.
So we have our LU decomposition,
and we know when it exists.
Before I end, two comments.
First, always check your work.
Always go back and
multiply L times U
and make sure it's
A, because it's
easy to screw up the
elimination process
and it's easy to
check your work.
So if you go back and make sure
things are as they should be.
Second comment is that
you might be worried
when you do this elimination
process that-- well OK, we
had to assume a is
non-zero because we
wanted a non-zero pivot.
You might worry
that we might have
to have a minus b be non-zero.
But in fact, a minus b can be 0.
It's not a problem
for this entry
to be 0 because we don't have
to do a row exchange to get
U. That's the only time when we
can't do the LU decomposition.
In particular, singular matrices
can have LU decompositions.
Good.
Thanks.
