In the last video we learned what it meant to take the
product of two matrices.
So if we have one matrix A, and it's an m by n matrix, and
then we have some other matrix B, let's say
that's an n by k matrix.
And we've defined the product of A and B to be equal to--
And actually before I define the product, let me just write
B out as just a collection of column vectors.
So we know that B can be written as a column vector b1,
another column vector b2, and all the way it's going to have
k of them because it has exactly k columns, so bk.
So the last video we defined the product of A and B , and
they have to have the columns of A have to be the same as
the rows of B in order for this to be well-defined.
But we define this product to be equal to A times each of
the column vectors of B.
So it's equal to-- let me switch back to that color--
it's equal to A times-- do it in the color that I did it-- A
times b1, and then the second column of our product is going
to be A times b2.
Our third product is going to be A times b3, all the way to
A times bk.
Right there.
And the whole motivation for this, you've probably you've
seen this before, maybe in your Algebra II class-- it
might have not been defined exactly this way, but this is
the equivalent to what you probably saw in your Algebra
II class-- but the neat thing about this definition is that
the motivation came from the composition of two linear
transformations whose transformation matrices were
the matrices A and B.
And I showed you that in the last video.
With that said, let's actually compute some matrix-matrix
products just so you get the hang of it.
So let's say that I have the matrix A.
Let's say that A is equal to the matrix 1, minus 1, 2, 0,
minus 2, and 1.
I keep the numbers low to keep our arithmetic fairly
straightforward.
And let's say that I have the matrix B, and let's say that
it is equal to 1, 0, 1, 1, 2, 0, 1, minus 1, and
then 3, 1, 0, 2.
So A is a 2 by 3 matrix, 2 rows, 3 columns.
And B is a 3 by 4 matrix.
So by our definition, what is the product AB
going to be equal to?
Well, we know it's well-defined because the
number of columns here is equal to the number of rows,
so we can actually take these matrix vector products--
you'll see that in a second-- so AB is equal to the matrix A
times the column vector, 1, 2, 3.
That's going to be the first column in our product matrix.
And the second one is going to be the matrix A times the
column 0, 0, 1.
The third column is going to be the matrix A times the
column vector 1, 1, 0.
And then the fourth column in our product vector is going to
be the matrix A times the column vector 1, minus 1, 2.
And this, when we write it like this, it should be clear
why this has to be, why the number of columns in A have to
be the number of rows in B, because the column vectors in
B are going to have the same number of components as the
number of rows in B, so all of the column vectors in B-- so
if we call this B1, B2, B3, B4, all of my bi's-- let me
write it this way-- all of my bi's where this i could be 1,
2, or 3, or 4, are all members of R3.
So we only have matrix vector products well-defined when the
number of columns in your matrix are equivalent to
essentially the dimensionality of your vectors.
That's why that number and that number
has to be the same.
Well, now we've reduced our matrix-matrix product problem
to just four different matrix vector product problems, so we
can just multiply these.
This is nothing new to us, so let's do it.
And so what is this equal to?
So AB,-- let me rewrite it-- AB, my product vector, is
going to be equal to-- so this first column is the matrix A
times the column vector 1, 2, 3.
And how did we define that?
Remember, one way to think about it is that this is equal
to the-- you can kind of think of it as the-- each of the
rows of A dotted with the column here of B, or even
better, this is the transpose of some matrix, right?
Let me write this this way.
If a is equal to-- sorry-- the transpose of some vector,
let's say that a is equal to the column vector 0, minus 1,
2, then a transpose-- and I haven't talked about
transposes a lot yet, but I think you get the idea.
You just changed all of the columns into rows-- so a
transpose will just be equal to 0, minus 1, 2.
You just go from a column vector to a row vector.
So if we called this thing here, a transpose, then when
we take the product of our matrix A times this vector, we
essentially are just taking A and dotting with this guy for
our first row and our first column.
So let me do it that way, so let me
write it in that notation.
So this is going to be the matrix-- or sorry-- the vector
1, minus 1, 2.
That's essentially, that row right there represented as a
column dotted with 1, 2, 3.
Actually, let me do it in that color just so I can later
switch to one color to make things simple-- but dotted
with 1, 2, 3.
So we just took that row, or, I guess the column equivalent
of that row, and dotted with this.
And I wrote it like this because we've only defined dot
products for column vectors.
I could do it maybe for row vectors, but we don't need to
make a new definition.
So that's going to be the first entry in this matrix
vector product.
The second entry is going to be the second row of A
essentially dotted with this vector right there, so it's
going to be equal to 0, minus 2, and 1, dotted with 1, 2, 3.
And we just keep doing that-- and I'll just switch maybe to
one neutral color now-- so that A times 0, 0, 1.
That's going to be the first row of A expressed as a column
vector, so we can write it like this, 1, minus
1, 2 dot 0, 0, 1.
And then, actually, and then we have our and then the
second row of A dotted with this column vector, so we have
0, minus 2, 1, dotted with 0, 0, 1.
Two more rows left.
This can get a little tedious and it's inevitable that I'll
probably make a careless mistake, but as long as you
understand the process, that's the important thing.
So the next one, this row of A expressed as a column vector,
1, minus 1, 2, and we're going to dot it with this vector
right there, 1, 1, 0.
And then this row of A-- I can just look over here as well--
0, minus 2, 1, dotted with 1, 1, 0.
And then finally the last two entries are going to be the
top row of A, 1, minus 1, 2, dotted with this column
vector, 1, minus 1, 2-- a little dot there, remember
we're taking the dot product-- and then finally this second
row of A, so 0, minus 2, 1, dotted with this column
vector, 1, minus 1, 2.
And that is going to be our product matrix-- and this
looks very complicated right now but now we just have to
compute it, and dot products tend to simplify things a good
bit-- so what is our matrix-- our product-- going to
simplify to?
I'll do it in pink.
AB is equal to-- draw the matrix right there.
So what is the dot product of these two things?
It's 1 times 1.
I'll just write it out.
It's 1 times 1, I'll just write, 1 times 1 is 1, plus
minus 1 times 2, so minus 2, plus 2 times 3, plus 6.
Now we'll do this term right here, 0 times 1 is 0, plus
minus 2 times 2, so that's minus 4, plus 1
times 3, plus 3.
Now we're on to this term, 1 times 0 is 0, plus minus 1
times 0, plus 0, plus 2 times 1 is equal to plus 2.
This term, 0 times 0 is 0, plus minus 2 times 0-- let me
write it 0 plus minus 2 times 0 is 0, plus 1 times 1.
So plus 1, and here you have 1 times 1 is 1, plus minus 1
times 1 is minus 1, plus 2 times 0, so plus 0.
Here 0 times 1 is 0, 2 minus 2 times 1 is minus 2, and then 1
times 0 is plus 0.
Almost done.
1 times 1 is 1; minus 1 times minus 1 is 1; 2 times 2 is 4.
Finally, 0 times 1 is 0; minus 2 times minus 1 is 2.
1 times 2 is also 2.
And we're in the home stretch, so now we just have to add up
these values.
So our dot product of the two matrices is equal to the 2 by
4 matrix, 1 minus 2 plus 6.
That's equal to 5.
Minus 4 plus 3 is minus 1.
This is just 2.
This is just 1.
Then we have 1 minus 1 plus 0 is just 0, minus 2, right?
We just have a minus 2 there, 1 plus 1 plus 4 is 6, and then
2 plus 2 is 4.
And we are done.
The product of A B is equal to this matrix right here.
Let me get my A and B back.
We can talk a little bit more about what this product
actually represented.
So let me copy and paste this.
Let me scroll down a little bit.
Go down here, paste.
There you go.
So this was our A and our B.
And when we took the product we got this matrix here.
Now there is another couple of interesting things to notice.
Remember, I only said that this product is only
well-defined when the number of columns in A is equal to
the number of rows in B.
So that was the case in this situation.
And then notice, we got a 2 by 4 matrix, which is the number
of rows in A times the number of columns in B.
So we got a 2 by 4 matrix.
So another natural question is, could we have found, or is
it even equal, if we were to take the product BA?
If we were to even take the product BA.
So if we tried to apply our definition there, what would
it be equal to?
It would be equal to the matrix B times the column 1,
0, then the matrix B times the column minus 1, minus 2.
And then it would be the matrix B times
the column 2, 1.
Now, can we take this matrix vector product?
We have a 3 by 4-- this right here is a 3 by 4 matrix, and
this guy right here is a member of R2.
So this is not well-defined.
We have more columns here than entries here, so we have never
defined a matrix vector product like this.
So not only is this not equal to this,
it's not even defined.
So it's not defined when you take a 3 by 4 matrix, and you
take the product of that with a 2 by 4 matrix.
It's not defined because that number and that
number is not equal.
And so obviously since this is defined and this isn't
defined, you know that AB is not always equal to BA.
In fact, it's not usually equal to BA, and sometimes
it's not even defined.
And the last point I want to make is, you probably learned
to do matrix-matrix products in Algebra II, but you didn't
have any motivation for what you were doing, but now we do
have a motivation.
Because when you're taking the product of A and B, we learned
in the last video that if we have two transformations,
let's say we have the transformation, S is a
transformation from R3 to R2, and that S is represented by
the matrix.
So S, given some matrix in R3, if you'd apply the
transformation S to it, it's equivalent to multiplying
that, or given any vector in R3, applying the
transformation S is equivalent to multiplying that
vector times A.
We can say that.
And I used R3 and R2 because the number of columns in A is
3, so it can apply to a three-dimensional vector.
And similarly, we can imagine B as being the matrix
transformation of some transformation, T, that is a
mapping from R4 to R3, where if you give it some vector x
in R4, it will produce-- you take the product of that with
B, and you're going to get some vector in R3.
Now, if we think of the composition of the two-- so
let's think about it a little bit-- if we have R4 here-- let
me switch colors-- we have R4 here, we have R3 here, and
then we have R2 here.
T is a transformation from R4 to R3.
So T would look like that.
T is a transformation and it's B times x.
That's what T is equal to, so T is this transformation.
And then S is a transformation from R3 to R2.
So S looks like that.
And S is equivalent to A times any vector in
R3, so that is S.
So now we know how to visualize, or how to think
about, what the product of A and B are.
The product of A and B is essentially, you apply the
transformation B first-- so let me think of it-- the
composition of S-- let me write it this way-- so what is
the composition of S with T?
This is equal to-- of x-- this is equal to S of T of x.
So you take a transformation from R4 to R3, and then you
take the S transformation from R3 to R2, so this is S of T.
S of T is a transformation from R4 all the way to R2.
And then the neat thing about this, if you were to just
write this out in its matrix representations-- we did this
in the last video-- this would be equal to the S matrix A
times this vector right here, which is Bx.
But now we know that the matrix-- by our definition of
matrix vector products-- that this guy right here is going
to have a transformation.
It's going to be equal to-- So the composition S of T of x is
going to be equal to the matrix AB, based on our
definition, so the transformation AB
times some vector x.
So the reason why I'm going all this is because we just
did a matrix-matrix product up here.
We took the pain of multiplying the matrix A times
the matrix B and we got this value here-- and hopefully I
didn't make any careless mistakes.
But the big idea here, the idea that you probably weren't
exposed to in your Algebra II class, is that this is the
matrix of the composition of the transformations S and T.
So right here, it's the matrix of the composition of S and T.
So you're not just blindly doing some-- matrix-matrix
products can be pretty tedious, but now you know what
they're for.
They're actually for the composition of two
transformations where each of A and B are the transformation
matrices for each of the individual linear
transformations.
Anyway, hopefully you found that useful.