Lesson 2.5: Arithmetic Part 1 (Old Version) - MATLAB Video Lecture | Study Introduction to Programming with MATLAB - IT & Software | Best Video for IT & Software

[MUSIC]
We're going to look at array addition.
There are special rules for
arithmetic on arrays.
Let's look at an example to illustrate it.
X and Y are two arrays that
we want to add together.
X is a 3 by 2 matrix, 1, 4, 7, 0, 5, 5.
And Y is also a 3 by 2 matrix,
2, -4, 6, 2, 0, and 3.
In order to add two matrices together,
we have to make sure that they
have the same dimensions.
So suppose we give MATLAB
the command Z equals X plus Y.
What's Z look like?
Well here's what X plus
Y equals to Z means.
On the right side, we see the two
matrices X and Y added together,
and on the left side, we see what happens.
Let's look at the first
element on the first row.
The first element on the first row of X and
Y is 1 and 2.
We add those together: 1 plus 2.
That's what we put at the first
element on the first row of Z.
Let's look at the next
element on that row.
X and Y are 4 and -4.
If we add those together,
we get 4 plus -4, which will be 0.
And we do this for
every position in the matrices.
If we carry out these individual additions,
we get 3, 0, 13, 2, 5 and
8 on the left side, and so
Z is equal to 3, 0, 13, 2, 5, and 8.
Let’s take a look at this in MATLAB.
We'll give MATLAB something to work with.
So let X equal 1, 5, -2
on the first row,
and then 3, 0, 7 on the second row.
And we'll be lazy with Y.
We'll use the colon operator.
So X is 2 by 3. That's six elements.
Y is 1 by 6, also six elements.
So let's add them and assign them to Z.
[CLICKING] Ah! MATLAB doesn't like this.
It says, “Matrix dimensions must agree.”
Well we've tried to add a 2 by
3 matrix and a 1 by 6 matrix,
and as we pointed out, that doesn't work.
So what I meant to do was
make Y a 2 by 3 matrix.
By putting a semicolon here
instead of a comma, and
now I'm going to use the up-arrow
to repeat that command.
Z equals X plus Y, and
things work just fine now.
Subtraction works great too.
Let's try that.
Let's let Z equal X minus Y.
And there's no problem there either.
Let's look at some of these numbers and
see if they came out right.
With addition, well let's look at
this second element on the first row.
We ended up with a 7.
Where'd that come from?
Well, there was a 5 at that position
in X and a 2 in Y.
5 plus 2 is 7.
And what about this -5 on the second
row of Z here, when we did subtraction?
Well, we had a 0 there for
X, and we had a 5 for a Y.
0 minus 5 gives us -5.
And we can do a similar thing
with array multiplication.
Array multiplication is very
similar to array addition.
We use the same example, X and Y,
that we used to illustrate addition, for
multiplication.
If we want to do array multiplication,
we use the dot asterisk operator.
So we say Z equals X dot asterisk Y.
You may wonder why we don't
just use the regular asterisk,
which is the traditional symbol for times.
That's because the asterisk alone is
reserved for matrix multiplication,
which is different from
array multiplication.
So what does Z look like?
Well, on the right side, you see the same
numbers you saw for array addition, but
this time the operator is the dot star
operator, and on the left the only
difference is: instead of plus everywhere
you see the asterisk, the times symbol.
So we have 1 times 2,
4 times -4, and so on.
So you carry out all these
multiplications, and
you can see the result for Z right here.
Well, array multiplication has the same
rule that addition and subtraction has,
that is that the matrices that are
operands have to have the same dimensions.
So we'll use the same examples, X and
Y, that we used for
addition and subtraction, and
here we go. Z is equal to X dot star Y.
That's array multiplication.
Really simple, and we can check an example.
Here's a 5 and a 2 on the first row
second element, and 5 times 2 is 10.
And if we look at the last
element on the second row,
we see a 7 and a 6, and 7 times 6 is 42.
So when might you use
array multiplication?
Well, let's clear the screen and
do an example.
Let's suppose you had an array of
temperature sensors, outdoors
somewhere, laid out in a rectangular grid,
covering a few square miles and
reporting back to a central facility.
And today, let's suppose they measured
these high temperatures which I've already
put into an array called Highs_measured.
Let's take a look at these. [CLICKING]
There.
So it's a 2 by 3 array.
You'd probably have a larger
array in a real application, but
this is just an example.
And furthermore, suppose that each
sensor's measured temperature must be
corrected by multiplying it by
the sensor's calibration factor.
And that would be kept in a permanent
array of the same shape. I've set up
that array too and called it
Calibration_factors.
Let's look at that one.
[CLICKING]
There.
And you can see both of these over in
the workspace over here on the right.
They were sitting there.
I don't know if you’d noticed them.
Well, to get the true temperatures,
we now need to multiply each element
in Highs_measured by the corresponding
element in Calibration_factors.
And that's exactly what array
multiplication does for us, so
let's do that.
Let's call the result Highs_true.
And we'll set that to
Highs_measured dot star
…There's that array
multiplication operator…
Calibration_factors, and there's the result.
If you look back at the original
temperatures up here, 71, 52,
78, and 83 and so on,
these are more reasonable down here,
78, 78, 77, and
just varying by a smaller amount.
They're more reasonable because
you wouldn't expect a wide
variation over just a few square miles.
The need for calibration like
this is fairly common, and
array multiplication makes it easy to do.
So now we know how array
multiplication works, and
we know that matrix multiplication is
very different from array multiplication.
In fact it differs in three ways.
First there's obviously no dot.
Second, the shapes of X and
Y must be compatible,
which is different from
having the same shape.
And third, the calculation of each
element of Z uses both multiplication
and addition.
But first let's look at
the compatibility requirement.
Here we show Z, X, and Y schematically.
And we've been careful to give
them all different shapes.
So X is clearly not the same shape as Y,
but it is as we'll see compatible with Y.
X and Y are compatible because the width
of X is the same as the height of Y.
We've highlighted that
with the blue arrows.
To be specific, the width of
the left operand of the matrix
multiplication operator must be equal
to the height of the right operand.
Put another way, the number of
columns of the left operand X
must equal the number of
rows of the right operand Y.
Let's move our schematic up a little
bit to make room to write their
dimensions, and there are X's dimensions.
And we've color-coded them to make it
easy to see that the number of rows,
L, is the height of X.
And the number of columns,
M, is its width.
Similarly, we show the dimensions of Y and
color-code them, so
it's obvious that the number of rows,
M, of Y is its height,
and the number of columns,
N, is its width.
If we drop the schematics and
put the dimensions under X and
Y in the matrix multiplication operation,
we notice that there are two
dimensions next to each other
that are equal, the two blue Ms.
Because of the way that these Ms
are situated between the L and the N,
they're called the “inner”
dimensions of X and Y.
And that provides a simple way to
state the compatibility rule for
matrix multiplication.
We simply say that the inner
dimensions of X and Y must be equal.
So, what about the L and the N?
Well, they're called the “outer” dimensions,
and there's
no restriction on them whatever, but
they do have an affect on the calculation.
They determine the shape of the result.
Z will be an L by N matrix.
So to review, the width of X,
blue, must equal the height of Y.
The height of Z, red,
equals the height of X.
And the width of Z, green,
equals the width of Y.
So, we know the shape of Z,
but we still don't know how
its elements are calculated.
Let's use MATLAB to see how that's done.
I've already prepared a couple of
matrices that we can multiply: A and B.
The dimensions are 4 by 3 for
A and 3 by 2 for B.
Since the width of A is
equal to the height of B,
we can do matrix
multiplication of A times B.
But before we do, let's put the sizes
into a row vector, like this.
Arranging their dimensions like this shows
that the inner dimensions are equal.
Here are the inner dimensions right here,
3, 3.
Which is another way to say that
the dimensions satisfy the rule for
matrix multiplication.
So let's multiple them and
assign the result to C.
So C is 4 by 2, and
those are the outer dimensions,
4 and 2, of the matrices A and
B, as expected.
Okay, so let's see how matrix
multiplication produced an element of C,
say the 0 on the third row.
That element is C(3,2), this one.
There, zero.
Since C(3,2) is on row 3 and column 2,
its calculation
involves only the elements on
row 3 of A and column 2 of B.
It’s calculated like this.
Now I'm going to write it out in detail.
There it is, and indeed it equals 0.
Now let's look at that expression.
Note that the row index stays the same for
A, so, 3, 3, 3.
And the column index stays the same for B.
So it's 2, 2, and 2.
But the inner index,
this one here, this 1 and
this 1, the inner index means it's
the inner index of these four,
and it's the same for the A and the B.
It ranges from 1 to 2 to 3.
It ranges across the entire width of A and
height of B.
The products of these pairs of elements
are then summed to get element 3, 2 of C.
This same sort of calculation is performed
for every row of A with every column of B.
And, and since there are four
rows of A and two columns of B,
we get a 4 by 2 matrix.
And what happens if
the inter dimensions do not match?
Well we can get an example of that by just
multiplying B and A in the other order.
MATLAB complains.
It says “Inner matrix
dimensions must agree.”
And to show that it's right to complain,
let's put the dimensions for
B times A into a row vector,
just as we did for A times B.
So, size B this time, size A.
Sure enough, the inner dimensions,
2 and 4, are not equal.
So MATLAB's right.
But then MATLAB's always right.
So that's how matrix
multiplication is carried out.
But when might you want to
use matrix multiplication?
Well, let's clear the screen.
There! Nice and tidy, and look at an application.
We'll use the same two matrices, A and
B, and here they are again, A, B.
Suppose a given a row of A represents
numbers of shares of each of three stocks.
And each row gives the shares
owned by one of four persons.
So person number 1 owns one
share of the first stock,
two shares of the second stock, and
three shares of the third stock.
Person 2 owns four, five, and
six shares respectively, and so on.
Suppose also
that each column of B represents
the gain in dollars of three stocks.
Column one gives the gains for
the stocks on one day, and
column two gives the gains
on the second day.
So looking at column one, we see that on
the first day the first stock gained $2,
the second stock gained $3 and
the third one gained $7.
Looking at column two, we see that on
the second day the first stock lost $2 and
the other two gained 8 and
$4 respectively.
Well to calculate the profit that one
person would have made on a given day,
we need to multiply the number
of shares times the gain for
each stock on that day and
then add the three products together.
So for example to calculate the profit for
the first person on the first day,
we would multiply one share times $2,
to get $2
and two shares times $3, to get $6
and three shares times $7 to get $21 and
then add these three products
together to get $29.
Then we'd repeat that process for
each person . . .
that's each row of A . . .
and each day . . . that's each column B.
Well that's exactly what
matrix multiplication does for us.
So let's do it.
[CLICKING].
There are four people in two days, so
there are eight distinct profits to
calculate, and the results are all
laid out neatly in matrix C.
Hm, let's see.
The first person profited about
the same on each day, $29 and $26.
The third person did much
better on the first day:
Didn't make anything on the second day,
and by comparing the columns, we can see
that everybody did better on the first
day than they did on the second day.
Matrix multiplication made
this calculation easy, but
this operator has applications far beyond
the stock market and far beyond finance.
The applications fall under
the heading of linear algebra, and
you can find many examples
in books on this subject.
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