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## Linear algebra

### Course: Linear algebra > Unit 2

Lesson 3: Transformations and matrix multiplication# Matrix product examples

Example of taking the product of two matrices. Created by Sal Khan.

## Want to join the conversation?

- How would I describe the solution set of a matrix geometrically where the set is:

x_1 = 1 + 2x_3

x_2 = 1 + 3x_3

x_3 = free variable

Thank you!(11 votes)- x_1 = 1 + 2t

x_2 = 1 + 3t

x_3 = t

which is the same as

(imagine that these are column vectors)

[x_1 x_2 x_3] = [1 1 0] + t[2 3 1]

So it's a line that passes through the point (1,1,0) and is parallel to the vector [2 3 1].(24 votes)

- So, when you see a 2x4 matrix, you can go, ha, that's a composite transformation from R4 to R2? And similarly, if you saw a 2x100 matirx, you could say, that's a transformation from R100 to R2? (Crossing dimensions :) ?(8 votes)
- Yes, that is correct. Any n by m matrix will multiply with a vector from ℝᵐ and create a vector from ℝⁿ.(6 votes)

- I heard about the idea that matrix multiplication can be visualized as applying one transformation on the entire coordinate plane after another, and I can visualize it quite well. I also understand that matrix product is non-commutative because when applying transformation on the entire coordinate, the order matters. However, I don't quite understand the "non-existent" case. Analytically, it looks obvious that matrix (m x n) x (l x m) is not possible. But when looking at it geometrically, I don't know how a transformation would not apply.

I am wondering if this is the case: the transformation does apply, but the combined transformation is no longer linear.

I hope this make sense... Thank you in advance!(3 votes)- I'm not sure what exactly a "geometric" meaning here would be; the fact of the matter is that it's just not defined. Let's say we have two matrices A & B with a being a m by n matrix and B being a p by q matrix. Now AB is only defined when n = p, it's outside the domain of the function otherwise. Realize that the geometric interpretation is exactly that - an interpretation. It is however, not the transformation itself. It describes the transformation. There are concepts in math that seem to intuitively make sense, but when you get your hands dirty they make less and less sense - that is there is no reasonable way for some properties, operations, or sets to exist. An example of this is the set of all possible sets. It's a fun exercise to think why this can't possibly exist, but on a base level seems like it should.(4 votes)

- If I have two transformations, T and S, and T goes from R4 to R3, S from R3 to R2 (as in the video at16:23), and T is represented by the matrix B, S by matrix A.

Sal multiplies AB(x). Why is it in that order? Considering it would - at least logically - make more sense to multiply BA(x).

I know we can't in this example, because BA is not defined. But I'm only interested in why we choose to use the matrix of the second transformation before the matrix in the first transformation.(3 votes)- Function composition is evaluated right-to-left because of how function notation is written. If I want to apply the function g, then the function f to the number x, I write f(g(x)).

Because matrices are (representations of) functions, the same applies here. Sal multiplied AB(x) because he wanted to apply the function T, then the function S. Multiplying BA doesn't make sense because outputs of A are 2-dimensional, and inputs of B are 4-dimensional.

Matrix multiplication is undefined precisely when the domains and ranges of the corresponding functions don't match up.(3 votes)

- (At the end, around17:20): It seems pretty clear that S(T(x)) = A*(Bx), because (e.g., if B is 3X4 and A is 2x3) Bx is an R3 vector and A*(Bx) transforms Bx; but have we proved this?(3 votes)
- It doesn't need to be proven because it is essentially the definition of matrix products. Matrix products (with appropriate sizes) just represent composition of functions.(3 votes)

- Hi there,

I just wana know if Sal talks about matrix sum notation in any of his videos. Like for example in my degree level physics we're doing a course on linear algebra, but the matrices in there are all about some weird indexes and sums...and proving stuff just by manipulating indices. I dont get it at all...is there any videos that I can watch?(3 votes)- Matrix addition and the commutativity of matrix addition appear in the first video: Introduction to matrices.(3 votes)

- Why is matrix multiplication not commutative?(3 votes)
- First of all, if
**A**and**B**are matrices such that the product**AB**is defined, the product**BA**need not be defined. In this case, matrix multiplication is not commutative. Secondly, if it is the case that both**AB**and**BA**are meaningful, they need not be the same matrix. For instance, let m and n be distinct, positive integers. Let**A**be an m by n matrix, and let**B**be an n by m matrix. Then the product**AB**is an m by m matrix, but the product**BA**is an n by n matrix. Since m and n are distinct,**AB**≠**BA**, and the operation is not commutative, etc. However, there are examples where this operation commutes, but this does not hold in general.(3 votes)

- What is the difference between sum of transformation of a particular vector and composition of transformation ?I know how each of them is computed but what is the real difference(2 votes)
- Well, I'm not sure what effect on the transformations direct addition would have. I doubt the result of the sum of transformation would have a direct relation to the individual transformations.

Linear transformation composition (multiplication) on the other hand is a way to join together multiple transformations into a single unit. The result is equal to the sequential application of the individual transformations.(3 votes)

- what does it mean to not being able to multiply
**B*****A**

in the example (of the video) why cant we have a transformation going from**R**4 to**R**2 (so it will be**B*****A**)?(2 votes)- We can absolutely have a transformation from R⁴ to R², but BA would be a transformation from R⁴ to R³, and then from R² to R³, which doesn't make sense.(3 votes)

- What does it mean to go from R3 to R2? Are we eliminating/ignoring a dimension?(2 votes)
- it means to transform the vector to 2 dimention by equaling the 3rd dimetion to 0.means now we will play with two dimention i.e x and y(if z eliminated)(2 votes)

## Video transcript

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.