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PROFESSOR: Hi there.

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My name is Ana.

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Welcome to recitation.

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In lecture, you've been learning
about how to multiply matrices,

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and how to think about
that multiplication

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in different ways, and
also about when a matrix is

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invertible or not, and how
to compute the inverse when

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it is invertible.

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And that's what today's
problem is about.

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We have a matrix A that
has variables a and b

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instead of numbers.

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And the question is: what are
the conditions on a and b that

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make that matrix invertible?

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And when it is invertible,
what is A inverse?

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Why don't you hit Pause and
work on it for a little while.

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And then we'll come back
and work on it together.

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And we're back.

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I hope you had success
in solving that.

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Let's do it ourselves.

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So, remember from
lecture, we talked

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about-- well,
Professor Strang talked

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about how it's easy to spot--
some easy tests to spot when

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a matrix is not invertible.

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Those were: if you have a column
of 0's or a row of 0's, then

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the matrix is not invertible.

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Or if you have two
columns that are the same,

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or two rows that are
the same, the matrix

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is also not invertible.

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So let's see if this
matrix satisfies

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any of those easy conditions.

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Well, here you
have a row of a's.

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If a is equal to 0, you have
a row of 0's, the matrix

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is not invertible.

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So a is-- What
was the other one?

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Two of the same column
or of the same row.

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Well, if a is equal to b, then
all the entries in the matrix

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are the same, so all
the rows are the same,

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all the columns are the same,
the matrix is not invertible.

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So that's the other
easy condition.

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A is not invertible if a is
equal to 0 or a equals to b.

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There's not any other easy
condition that I can spot.

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So we have to do it in
a more systematic way.

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So to do that, actually what
we do is we start with A,

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we try to find its inverse,
and if in the process

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we run into something fishy,
those are our conditions.

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So remember how to find
the inverse of a matrix?

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You start by writing a
giant matrix that has

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A and the identity next to it.

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And then you perform
elimination steps

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until-- let's put dots here,
because there's lots of steps--

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and you stop once you've
reached the identity

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matrix on the left side.

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And when you do that, what
you have on the right side

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will be your A inverse.

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Easy enough.

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So let's do the computations.

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a, b, b; a, a, b; a, a, a; and
my identity matrix next to it.

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And now I know you
learned about elimination

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in the past recitation.

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So I'll do the first
few steps slowly,

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and then I'll just write the
solution for the last steps.

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So we want to eliminate this a.

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We want to turn it into 0.

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So let's subtract the
first row from the second.

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Let's do row 2 minus row 1,
and write it instead of row 2.

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So a, b, b, 1, 0,
0; 0, so a minus a,

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a minus b, b minus b is 0, 0
minus 1, 1 minus 0, 0 minus 0.

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And then we want to do the
same with the third row.

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So let's do row 3 minus row 1.

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0, a minus b, it doesn't fit.

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a minus b.

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Can you understand that?

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And then 0 minus 1,
0 minus 0, 1 minus 0.

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All right.

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Let's continue up here.

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Now I want to eliminate
this a minus b.

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I want to turn it into a 0.

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So the first row stays the same.

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

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And now 0 minus 0, a minus
b minus a minus b is 0.

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a minus b minus 0, that's easy.

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-1 minus -1, 0 again.

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-1 and 1.

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Almost there.

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We already have 0's down here.

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It's looking more and more
like the identity matrix.

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So we have to turn all these
diagonal entries into 1's, so

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let's do that now.

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1, b over a, b over-- oh.

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I'm dividing by a.

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a better not be 0.

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Let me signal that here.

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a different from 0.

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1 over a, 0, 0.

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0.

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I want to make that into a 1.

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So 1 over a minus b, and--
you guessed it-- a minus

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b has to be different from 0.

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0.

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-1 over a minus b.

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1 over a minus b.

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0, 0, 1, 0, -1 over a minus b.

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1 over a minus b.

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AUDIENCE: [INAUDIBLE]

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PROFESSOR: Someone's
pointing a mistake.

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AUDIENCE: Yeah, the (2, 2)
entry, you should have a 1.

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PROFESSOR: Oh, you're
absolutely right.

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Thank you.

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All right.

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So not much left to do.

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We only have to eliminate
b over a and b over a.

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That's a little bit
computationally heavy.

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So what do you have to do?

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You just have to
basically subtract--

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replace row 1 by row 1 minus b
over a times row 2 plus row 3.

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Let me write that down.

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row 1 minus b over a
times row 2 plus row 3.

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You want that to
go into your row 1.

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These stay the same.

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-1 over a minus b.

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1 over a minus b.

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This is where I'm
going to go to my notes

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and tell you that these
numbers are 1 over a minus

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b, 0 minus b over a, a minus b.

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That looks pretty awful.

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But that is A inverse.

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And we're done.

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Except it looks terrible.

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So let me just write
it in a nicer way.

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A inverse equals-- see,
you're dividing everything

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by a minus b, so
pull out that factor.

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

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Looks much better now.

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So here's your A inverse.

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And the other question was: what
are the conditions on a and b

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for the matrix A
to be invertible?

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Well, a has to be non-zero, and
a has to be different from b.

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And we're done.

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That's it for today.

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See you next time.