WEBVTT

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

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going to delve a little bit more
deeply into matrix

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multiplication and look at some
of the properties and the

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conditions under which different
sorts of multiplication can be

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

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

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

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

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it's three by three.

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

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So it's a 3 by 2.

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And we notice that these
numbers are the same.

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

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

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

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

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

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

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

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In the product.

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Let's try and work it out.

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

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

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the Second Matrix.

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What we want is 3
* 1 which is 3.

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

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ones or three.

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

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paired elements together and
add the result.

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

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first row here and the second

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

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2 * 3.

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Is 6.

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1 * -- 4 is minus 4.

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

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together and adding the results.

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

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

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

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Minus 3 * -- 2 is +6.

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2 * 3 is 6.

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

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

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column. 4 * -- 2 is minus 8.

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Minus 3 * + 3 is minus 9.

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2 * -- 4 is minus 8.

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

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

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

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4 * -- 2 is minus 8.

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3 * 3 is 9.

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

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which is minus 10.

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

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

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subtract 4, which is 2.

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Minus 6 + 6 zero subtract 4 is

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minus 4. Four and
six is 10 and 616.

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

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5 subtract 8 + 9.

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6th

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and minus 10 + 12 -- 12 is minus

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10. And this is the result
of multiplying these two

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matrices together.

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What about if we try and
multiply the two

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matrices together the
opposite way round?

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Suppose we try and
workout N * M.

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

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

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

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And what we observe now is
that these two numbers here

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

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cannot do the matrix
multiplication in the order

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

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

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that even when you can find a
matrix product by multiplying

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two matrices together, it
matters very much. The order

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

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

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some more examples.

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

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

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

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either of these exist.

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

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And in the second example
here, within the Second matrix

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

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work this product out because
these numbers are the same and

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

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the answer will be 2 rows and
two columns.

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

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

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

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It's a two by three matrix.

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

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

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

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

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

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

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

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to 3 * -- 1, which is 10.

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And you can check for yourself
that the remaining elements are

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131 and minus 11.

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

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two by two matrix.

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

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

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

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

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

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

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

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9. 1521
789

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

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multiply C * D together.

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

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get may have completely
different sizes. It's certainly

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

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

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

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we say that matrix
multiplication is not

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

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in which you carry out the

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multiplication. Now that we know
how to multiply 2 matrices

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

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identity matrices. Suppose we
have a two by two identity

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matrix, that's 1001.

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

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And suppose I want to multiply
these two together.

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The identity matrix is
certainly a two by two matrix,

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

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these numbers are the same, we
can actually workout the

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

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

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

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

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

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4. Which is just 1 * 2 is 2.

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

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* 7, which is just three.

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

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

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so we just get minus 4.

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And finally, the last element is

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

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

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identical to the matrix we
started with here. In other

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words, multiplying a matrix by
an identity matrix when this

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multiplication is possible
leaves an answer which is

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

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important property of identity

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

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

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

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

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

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Two times nothing
is nothing 313.

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

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

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

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

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

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matrix unaltered, identical to
what it was before.

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The same works even if we
haven't got square

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matrices. Suppose we have
this identity matrix.

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And we multiply, for example by
the Matrix 78.

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

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

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the matrix multiplication and
the result is going to be a

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

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started with.

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

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

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the result is just 7th.

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

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

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

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message that multiplying by an
identity matrix leaves the

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original matrix unaltered.