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BEN HARRIS: Hi.

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I'm Ben.

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Today we are going to do an
LU decomposition problem.

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Here's the problem right here.

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Find that LU decomposition
of this matrix A.

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Now notice that this matrix
A has variables, as well

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as numbers.

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So the sentence
ends: when it exists.

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And the second part
of the question

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asks you: for which
real numbers a and b

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does the LU decomposition of
this matrix actually exist?

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Now, you can hit pause now and
I'll give you a few seconds.

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You can try to solve
this on your own,

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and then we'll be back
and we can do it together.

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And we're back.

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Now, what do you have to
remember when doing an LU

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decomposition problem?

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Well, we do elimination
in the same way

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that we did before in order to
find U. But with this question

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we need to find L as well.

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So we need to do
elimination, but we also

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need to keep track of
the elimination matrices

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along the way.

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Good.

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So let's do that.

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So let me put my matrix up here.

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And we want to do elimination.

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So which entry do
we eliminate first?

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That's right.

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It's this (2, 1) entry.

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So we replace the second
row by the second row

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minus a times the first
row, and we get this.

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But we're not just
doing elimination,

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we're finding an
LU decomposition.

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So we need to keep
track of the matrix

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that I multiplied the
elimination matrix, that I

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multiplied this matrix A by on
the left to get this matrix.

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So what is that?

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That's this E_(2,1).

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Since I eliminated the (2, 1)
entry, I'll call it E_(2,1).

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And it's this matrix.

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Why is it this matrix?

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Well, remember how
multiplication on the left

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works.

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I replaced the first row
by just the first row.

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I replaced the second
row by the second row

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minus a times the first row.

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So you can just read
off from these rows

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which operations I did.

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Now, which entries
should we eliminate next?

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We need to eliminate this b.

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So we will replace the
third row by the third row

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minus b times the first row.

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And which elimination
matrix did we use?

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Well, note, we
replaced the third row

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by the third row minus
b times the first row.

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That's exactly what you should
read off this elimination

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matrix.

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Good.

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Now, we only have one step left.

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We only need to
eliminate one last entry.

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But this one's a little
tricky, so let's be careful.

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In order to eliminate this
b, we need a to be a pivot.

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In particular, we
need a to be nonzero.

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If a were zero
here, then we would

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have to do a row exchange.

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And that's no good.

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You can't find an
LU decomposition

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if you have to do a row
exchange in elimination.

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So we need to assume
that a is non-zero

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in order to keep going.

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So let's just assume
there that a is non-zero.

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Now, what do we do?

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Well we can replace
the third row

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by the third row minus b
over a times the second row.

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And we just get this.

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a minus b.

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And let's write down
our elimination matrix.

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E_(3,2) now.

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There's our elimination matrix.

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We replaced the third row
by the third row minus b

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over a times the second row.

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Good.

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So we found our U matrix.

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That's what elimination does,
it gives us our U matrix.

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So let me write it up here.

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1, 0, 1; 0, a, 0;
0, 0, a minus b.

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Good.

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Now we have to
find our L matrix,

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and we need to use these
elimination matrices

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that we've been recording along
the way in order to do that.

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So remember that we started
with A, and we got U.

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And how did we do that?

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Well we multiplied on the left
by all of these elimination

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matrices, E_(2,1),
E_(3,1) and E_(3,2).

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Sorry if that's
scrunching together there.

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Now, if we move these
elimination matrices

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to the other side then we'll
get L. So what do we have?

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We have A equals E_(2,1)
inverse, E_(3,1) inverse,

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E_(3,2) inverse times
U. And this is our L,

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this product of
these three matrices.

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Good.

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So let's compute it now.

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So L is the product
of three matrices.

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I need to get them by going
back and looking at these three

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elimination matrices and
taking their inverses.

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Well the nice thing
about taking an inverse

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of an elementary
matrix like this is we

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just make a minus a
plus or a plus a minus.

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So that's easy enough.

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We just change the
off-diagonal entries,

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we just change their signs.

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You can check that that
does what we wanted it to.

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It gives us the inverse.

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Good.

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And the last comment is
that multiplying these three

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matrices is really
easy in this order.

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Turns out all you do is you just
plop these entries right in.

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1, 1.

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Good.

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So this is our L matrix.

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So now we have our U
matrix and our L matrix,

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and we're done with the
first part of the question.

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The second part asks us for
which real numbers a and b

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does this decomposition exist?

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Now let's go back and
remember that at one point

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we had to assume
that A was non-zero.

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That was the only
assumption we had to make

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to get this decomposition.

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So it exists-- it being
this decomposition--

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when a is non-zero.

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And that's the answer
to the second part.

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So we have our LU decomposition,
and we know when it exists.

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Before I end, two comments.

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First, always check your work.

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Always go back and
multiply L times U

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and make sure it's
A, because it's

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easy to screw up the
elimination process

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and it's easy to
check your work.

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So if you go back and make sure
things are as they should be.

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Second comment is that
you might be worried

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when you do this elimination
process that-- well OK, we

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had to assume a is
non-zero because we

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wanted a non-zero pivot.

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You might worry
that we might have

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to have a minus b be non-zero.

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But in fact, a minus b can be 0.

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It's not a problem
for this entry

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to be 0 because we don't have
to do a row exchange to get

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U. That's the only time when we
can't do the LU decomposition.

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In particular, singular matrices
can have LU decompositions.

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Good.

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Thanks.