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CHRISTINE BREINER: Welcome
back to recitation.

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In this video, I
really just want

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to practice matrix
multiplication, which

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is potentially something
new for some of you,

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and maybe some of you have
been doing it for a while

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and are very good at it.

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But I want to make sure that
everyone is feeling confident

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in their ability to
multiply matrices.

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So we have three matrices here.

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We have A, B, and C. And
what I want you to do

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is I want you to compute
what makes sense below.

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I have four products
of matrices below.

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a is A times B, b is B
times A, c is B times C,

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and d is A times
C. So I want you

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to multiply the matrices
that make sense to multiply,

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and then the ones that don't,
make sure you understand why.

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Give yourself a
brief explanation

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of why you can't multiply them.

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So why don't you work on
that, pause the video,

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and when you feel
confident in your answers,

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bring the video back up, and you
can check them against my work.

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OK, welcome back.

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Well, we wanted to make sure
we felt comfortable multiplying

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matrices.

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So what we're going to do
is look at the four products

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I mentioned below, and we're
going to see how they do,

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whether or not we can
actually compute them.

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So let's first look
at a, which was

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A times B. So before
I write it down again,

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A times B is-- notice A is a
2 by 2 matrix, and B is a 2

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by 3 matrix, right?

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And so if I write letter
a, we know we're taking a 2

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by 2 by a 2 by 3.

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And so the fact that the
interior dimensions agree,

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that the number
of columns of A is

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equal to the number
of rows of B,

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means that I can multiply them.

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So I can multiply them, and
my result I expect to get

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is, of course, the dimensions
we have on the outside.

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So I expect to get
a 2 by 3 matrix.

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So I'm going to
rewrite A and B here,

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so that I don't have to
keep walking back and forth,

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and then we'll do
the multiplication.

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So I have [6, 5; 1, 2]
times [2,  -1, 3; 1, 0, 4].

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OK.

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So I want to perform
this multiplication.

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Now remember that
when you are looking

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for a value in your resulting
matrix, which I know is 2 by 3,

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so I can even make a
little space for myself.

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I know it's 2 by 3,
so I know I'm going

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to have to fill in these spots.

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When I look at this position,
it's row 1, column 1.

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That means I take row
1 of the first matrix,

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and I'm essentially
just dotting it

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with column 1 of
the second matrix.

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So I'm taking row 1 times column
1 in the way it was described,

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which is I take 6 times 2,
and I add it to 5 times 1.

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So row 1, column 1 gives me
6 times 2 is 12, plus 5 times

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1 which is 5, so I get 17.

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And then if I come in to
the next spot, what is this?

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And the resulting matrix's
position is row 1, column 2.

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So now I take row 1
of the first matrix

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and column 2 of
the second matrix,

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and I get 6 times negative 1
is negative 6, plus 0 times 5,

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so I get a negative 6 here.

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Negative 6 times 0.

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Maybe I should
show you this way.

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Negative 6 times 0.

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OK.

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And then here I am now in the
third spot of the first row,

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so I'm in a row 1, column 3.

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So that's again, row 1 of the
first, column 3 of the second.

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So you see a pattern here about
where we're getting our things

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from that we're multiplying.

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For row 1, column
3 of the resulting,

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I take row 1 of the first
and column 3 of the second.

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So 6 times 3 is 18,
plus-- 5 times 4 is 20.

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So 20 plus 18 is 38.

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OK.

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Now I have to do the
same thing on the bottom.

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Right?

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So I have now here, the
row, notice I'm in row 2,

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so I'm always going to use
row 2 of this first matrix.

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And then what we
saw last time is I

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used column 1 in the
first spot, column 2

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in the second spot, column
3 in the third spot.

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Right?

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That's what happens
over and over again.

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So what do I do?

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I take [1, 2], and I
multiply it by [2, 1].

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So I take 1 times
2, plus 2 times 1.

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So I get 2 there and
2 there, so I get a 4.

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And then the next
column: row 2, column 2.

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1 times negative 1 is
negative 1-- 2 times 0

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is 0-- so I get a negative 1.

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And then the last column.

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1 times 3 is 3.

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2 times 4 is 8.

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So I get 3 plus 8, so I get 11.

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Hopefully I didn't make any
stupid summing mistakes there,

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but if I did, you
probably caught it.

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Because I was trying to
say what I did as we went.

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So that is the answer to a.

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OK.

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So now let's think
about what is b.

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b was take B times A, which
is just to switch the order.

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So let's look at the
dimension match-up.

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Now we have a 2 by
3 matrix, and I'm

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trying to multiply it
by a 2 by 2 matrix.

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Well, I don't have
to do any more work,

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because I can't do it.

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Because the dimensions of the
insides here-- three columns

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for B, two rows for A--
means that I can't actually

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multiply them.

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OK?

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So this isn't even defined.

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OK, so that was easy.

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That was b.

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All right.

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Letter c-- I'll give
myself a lot of room

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to do that-- letter
c was B times C.

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And so I'm going to
write down the dimensions

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to see if I even need to
write down the matrices.

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B was two rows by
three columns, and C

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was three rows by two columns.

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So if I look at the dimensions,
the 3 and the 3 match up,

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so I am going to be
able to multiply them,

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and my result-- as I mentioned
before-- should be a 2 by 2.

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So let me write
down B and C here,

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so we don't have to
keep going to the side.

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OK.

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And then C is 1, 2 negative
1; negative 1, 3, 2.

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All right, let me just make sure
I didn't transcribe anything

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incorrectly.

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I think it looks good.

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OK.

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So row 1 of B. Row 2 of B.
Column 1 of C. Column 2 of C.

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We're going to be dealing
with those, specifically.

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So we want to multiply these.

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We said our resulting matrix
is going to be 2 by 2.

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OK.

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Because I'm going to
have a lot of terms,

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I might write them down on
this one, and then simplify.

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Because I may make a mistake,
so to be more careful,

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I'll write down all the pieces.

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So here I am in row 1,
column 1 of the resulting.

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So I take row 1 of the first,
column 1 of the second,

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and what do I get
when I multiply?

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I get 2 times 1-- that's 2.

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Plus negative 1 times
2-- that's negative 2.

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Plus 3 times negative
1-- that's negative 3.

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Right?

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That's all we have to do
for the first position.

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Then I do, for the second
one, it's row 1, column 2.

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So I do row 1, column 2.

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So I'll try to keep my
head out of the way.

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I realize I keep
stepping in front.

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So it's 2 times negative 1.

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I get negative 2.

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Negative 1 times 3,
so I get negative 3.

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And 3 times 2 gives me 6.

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OK.

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And then the bottom two.

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I get row 2, column
1 over here, and then

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row 2, column 2 over here.

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So row 2, column 1 is
going to be 1 times 1.

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Plus 0 times 2.

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Plus 4 times negative
1, so I get negative 4.

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And then here.

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Row 2, column 2.

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I get 1 times negative
1, so I have negative 1.

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Plus 0 times 3-- plus 0.

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And then 4 times 2 is 8.

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So if I simplify these, it
looks like in the first spot

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I should get a negative 3.

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And the second spot,
I should get a 1.

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This is just for you
to check your answer.

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And the third spot,
I get a negative 3.

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And then the fourth
spot, I get a 7.

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So hopefully I added
correctly all throughout.

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I think I did, so I
think we're good there.

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So that is the answer to c.

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And again, the reason
we can multiply those,

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was that the dimensions-- when
you wrote them down in order--

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the dimensions to
the inside agreed,

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and then the outside
gives us the size

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of the resulting matrix.

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So there was one more
problem, and that was d.

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And I wanted you to take A
times C. And A was a 2 by 2.

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And C was a 3 by 2.

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And so again, we
see we can't do it,

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because the two
interior dimensions

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here-- when I write them in
that order-- don't agree.

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OK.

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So d is not defined.

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All right, so the basic
idea of this whole video

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is just to make sure we
felt comfortable multiplying

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matrices.

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We're trying to use some simple
examples to understand that.

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Understand how we can
recognize from the dimensions

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whether or not multiplication
is even defined,

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and then what size the
resulting matrix will be.

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I do want to point
out one thing.

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And I want to point out that
if we come over to our example

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back in the beginning.

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We had A*B as our first example
and then B*A as our second

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example.

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And A*B-- well, I think, I got
to remember what they were--

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yeah, A*B you could multiply,
but B*A you could not.

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So I think it has
been stressed before,

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but I think I should
stress it again, that order

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matters in multiplication.

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OK?

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You can't commute these things.

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You can't switch the order
and get the same result.

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So matrix
multiplication, you have

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to be very careful about
keeping things in the same order

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as you're multiplying.

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OK, I think that is
where I will stop.