https:/.../matrixmultiplication2f61mb-aspect.mp4

In this second of the videos on
matrix multiplication, we're

going to delve a little bit more
deeply into matrix

multiplication and look at some
of the properties and the

conditions under which different
sorts of multiplication can be

carried out. Let's start by
looking at looking at a specific

example in this example. Here
I've written down to matrices

M&N. And let's look at the sizes
of these matrices. The first

matrix M. Is a three row
three column matrix, so

it's three by three.

And the second matrix N is 3
rows, two columns.

So it's a 3 by 2.

And we notice that these
numbers are the same.

The number of columns in the
first is the same as the number

of rows in the second, so we can
perform this matrix

multiplication and the size of
the answer will be a three by

two matrix. So right at the
start we know the size of the

answer. It's going to have three
rows and two columns just like

this one had. So the shape
of the answer is.

Like we have here and we're
looking for these 6 numbers.

In the product.

Let's try and work it out.

To find the number that's in the
first row, first column. We work

with the first row of the first
matrix and the first column of

the Second Matrix.

What we want is 3
* 1 which is 3.

2 * -- 2 which is minus 4 and 1
* 3, which is 3. So we've got 3

ones or three.

2 * -- 2 is minus 4 and 1
* 3 is 3. We multiply the

paired elements together and
add the result.

When we come to the first row,
second column, we work with the

first row here and the second

column here. And again, pairing
off 3 * -- 2 is minus 6.

2 * 3.

Is 6.

1 * -- 4 is minus 4.

So in each case, we're
multiplying the paired elements

together and adding the results.

When we want the element that's
going in here, which is in the

2nd row first column of the
answer, we work with the 2nd

row, first column of the given
matrices 4 * 1 is 4.

Minus 3 * -- 2 is +6.

2 * 3 is 6.

And continuing in the same way,
the answer that goes in the 2nd

row, second column comes from
taking the 2nd row, second

column. 4 * -- 2 is minus 8.

Minus 3 * + 3 is minus 9.

2 * -- 4 is minus 8.

And finally on the last row
to find the element in the

first row. Sorry the 3rd row
first column will work with

the 3rd row, First Column, 5
ones of five.

4 * -- 2 is minus 8.

3 * 3 is 9.

And similarly to find the last
element, it will be 5 * -- 2,

which is minus 10.

4 * 3 is 12 and 3 * --
4 is minus 12.

And if we just tidy up what
we've got, we'll have 336

subtract 4, which is 2.

Minus 6 + 6 zero subtract 4 is

minus 4. Four and
six is 10 and 616.

Minus 8 -- 9 --
8 is minus 25.

5 subtract 8 + 9.

6th

and minus 10 + 12 -- 12 is minus

10. And this is the result
of multiplying these two

matrices together.

What about if we try and
multiply the two

matrices together the
opposite way round?

Suppose we try and
workout N * M.

Now, in this case the size of
the first matrix here is 3 rows

and two columns, so that's a
three by two and the size of the

second matrix is 3 by 3, three
rows, three columns.

And what we observe now is
that these two numbers here

are not the same, they are not
equal. That means that we

cannot do the matrix
multiplication in the order

that I've written it down
here. That matrix product

doesn't exist. So this is the
first point. I'd like to make

that even when you can find a
matrix product by multiplying

two matrices together, it
matters very much. The order

in which you write them down.
It may be possible to workout

a product one way, but not
another way. Let's look at

some more examples.

Suppose we've got two matrices
C&D as I've written them down

here, I'm going to try to work
out the product C * D.

And I'll also try and workout
the product D times. See if

either of these exist.

But in the first case, we've got
a two row three column matrix.

And in the second example
here, within the Second matrix

here we've got three rows into
two columns, so we can in fact

work this product out because
these numbers are the same and

the result will be a two by
two matrix. So the shape of

the answer will be 2 rows and
two columns.

If we try and do this the other
way round, D * C, The first

matrix Now has got three rows
and two columns. It's a three by

two matrix and the second one's
got two rows and three columns.

It's a two by three matrix.

So you can. You can see that we
can still work it out because

these two numbers are still the
same 2 into the same, but this

time the result is going to be a
three by three matrix, so it's

going to be a bigger matrix with
three rows and three columns.

We can use the process that we
evaluate that we worked on

before to evaluate the elements
in the these matrices. So for

example, the element that goes
in here is 1 * 3 + 2 * 5 added

to 3 * -- 1, which is 10.

And you can check for yourself
that the remaining elements are

131 and minus 11.

So it's possible to workout
C * D and the answer is a

two by two matrix.

When we do it the other way
round, let's take an element

here. Let's take the elements in
the first row, first column and

we obtain the answer by working
with the first row, first

column. Here, that's three
times, one is 3 added to minus 7

* 4. That's three added to minus
28, which is minus 25.

And you can proceed in the same
way to fill out this resulting

matrix and the numbers. You'll
get a -- 25 -- 29 -- 33.

9. 1521
789

The important point that I want
to make here is that when you

multiply C * D together.

It may be possible to also find
D * C, But the answers that you

get may have completely
different sizes. It's certainly

not true that CD is the same as
DC, so one of the observations

we take away straight away is
that in general CD is not equal

to DC. Even in situations where
both of these products do exist,

we say that matrix
multiplication is not

commutative. In general, it
really doesn't matter the order

in which you carry out the

multiplication. Now that we know
how to multiply 2 matrices

together, I'm going to show you
an important property of

identity matrices. Suppose we
have a two by two identity

matrix, that's 1001.

And suppose we have a second
matrix, two 3 -- 4 and seven.

And suppose I want to multiply
these two together.

The identity matrix is
certainly a two by two matrix,

and this matrix is also a two
by two matrix. So because

these numbers are the same, we
can actually workout the

product and the answer is also
a two by two matrix. So the

answer has this sort of shape
with four elements in there.

To get the first element in the
answer, we want to pair 10 with

2 -- 4, multiply the paired
elements together and add so we

get 1 * 2 is 2 added to 0 * --

4. Which is just 1 * 2 is 2.

To get this element here, we
want 1 * 3 which is 3 added to 0

* 7, which is just three.

To get the element in here, we
want to pair 01 with two and

minus four, so it's 0 * 2, which
is nothing 1 * -- 4 is minus 4,

so we just get minus 4.

And finally, the last element is

0 times. Three, which is nothing
1 * 7 is 7, so that's our

answer. And if you look at the
answer you'll see the answer is

identical to the matrix we
started with here. In other

words, multiplying a matrix by
an identity matrix when this

multiplication is possible
leaves an answer which is

identical to the matrix you
started with, and that's a very

important property of identity

matrices. The same result occurs
if we do the multiplication the

other way round. If we take two
3 -- 4 seven and we multiply it

by the identity matrix, one
nought nought one will find.

It's also possible, and if you
go through the operation 2 * 1

is 2 three times. Nothing is
nothing. The result there is 2.

Two times nothing
is nothing 313.

Minus 4 * 1 added to 7 times
nought is minus 4.

And minus four times North,
which is nothing added to 717

and you'll see again this answer
here is the same as this matrix

here. So that's very important
property to remember when you

multiply a matrix by an identity
matrix, it leaves the original

matrix unaltered, identical to
what it was before.

The same works even if we
haven't got square

matrices. Suppose we have
this identity matrix.

And we multiply, for example by
the Matrix 78.

Well, this has got one row and
two columns. It's a one by two

matrix. This is got two rows,
two columns, so we can perform

the matrix multiplication and
the result is going to be a

one by two matrix that's the
same shape as the one we

started with.

And if we carry out the
operations, it's 7 * 1, which

is 7 added to 8 times
nothing, which is nothing. So

the result is just 7th.

7 times and nothing is nothing
and 8 * 1 is 8, so it's just

eight. And again, this answer 7
eight is the same as the matrix

we started with over here. So
that's just to reinforce the

message that multiplying by an
identity matrix leaves the

original matrix unaltered.