nananana... multiplication of matrices. Matrix A times matrix B is possible if the number
of columns in A equals the number of rows in matrix B. Or, in symbols matrix A times
matrix B, A having the order of m by n and B having the order of n by p, again the columns
in A has to be equal to the number of rows in B. Now when you get multiplying these two
matrices, I was going to say binomials, these matrices together that new matrix, the product,
is going to have the order which is equal the number of rows in A that first matrix
in your multiplication by the number of columns in B that second matrix. Now, this might give
you an idea that with all of these restrictions about how these columns have to equal these
rows and how this is going to be the size of the new matrix, that you can't just turn
around the matrices, or flip-flop, or move...change the order of the multiplication of A to B
and expect to get the same answer. Indeed, you will see this in the properties scene
as we end this video, but multiplication is not commutative with matrices like it would
be with just addition, excuse me with just generic multiplication of numbers. The definition
of matrix multiplication. The product of an m by n matrix A and an n by p matrix B is
now going to be a new m (rows from the first matrix) by p (the columns of the second matrix)
AB. The elements in the i th row and the j th column of AB is found by multiplying each
element in the i th row of A by the corresponding element in the j th column of B and then adding
up all the products however many there may be. You will see this, especially it will
really show up on the last example. The first three are going to be of a, a little bit of
a special case. We are going to match the corresponding elements the row of the first
matrix to the column of the second matrix. So, I can look at these words all I want and
point over here, but it probably is not going to make total sense until you see it worked
out. So, let's do that. Here is our first three examples. We have got matrix A is a
one by three matrix of three, negative two, five. Matrix B is again a one by three matrix
with negative one, four, seven. Finally C is a one column three row matrix or a one
by three. So, we have, excuse three by one. So, we have a one by three, a one by three,
and a three...one, two, three rows...by one column. I guess that I could put that up there...
the right way. Ok, so the first thing that we are going to do, is we are going to take
matrix A and multiply it by matrix B. There is A, there is b, they are both one by three
matrices, they have the same order so we should be able to multiply them right? No, that is
good for when you are adding or subtracting matrices, not when you are multiplying. We
are tying to take a one by three matrix and multiply it by a one by three matrix. Well,
if you have to match all of the elements in the i th row of A with all of the corresponding
elements in the j th column in matrix B, then the number of columns in the first matrix
has to match up with the number of rows in the second matrix. That would be something
like me trying to do like three times negative one which is good. But, then when I get into
my...as I move through the row of my first matrix and I see negative two that needs to
correspond with some number in this first column. It would be first row and first column,
and that is good but then you know that is not going to be good. You don't go across
the rows. You go across the row of the first matrix by the first column of the second.
This is not going to have an answer. There is no solution. Now, in this part b we are
going to take matrix A and multiply it by matrix C. Well, A is a one by three matrix
as we can see. C is a three by one matrix. And of course I covered up the symbol type
thing, because I am really going to pay attention to this second part of the definition of matrix
multiplication here. But, we do have a number of columns in the first matrix equal to the
number of rows in the second matrix. This is going to be the number of rows in the first
matrix and the number of columns in the second matrix are going to be our new order. When
we get done doing matrix A times matrix C, we are just going to have a one by one matrix.
So, that is going to look like, found by multiplying each element in the i th row of the first
matrix matrix by the corresponding element in the j th column of B or the second matrix and then add up all of the products.
A is first and C is second, so row and column. First element in the first row, first element
in the first column. We have three times negative two. Then the second element in the first
row and the second element in this first column, negative two times zero. Now plus the third
element of the first row and the third element in the first column and we have five times
one. That is going to be negative six plus zero plus five. We have our one by one matrix
which is negative six plus five is negative one. Just checking my notes here, excellent.
We have the product of matrix A times matrix C is simply negative one. Now we are going
to take matrix C and multiply it by A. Now again, we are going to put matrix C as the
first matrix and matrix A as the second. We are going to find that multiplication by multiplying
each element in the first matrix, the row of the first matrix by the corresponding element
in the column of the second matrix and add up all of those products. So let's see here.
C first then A. So, first row first column, or you know the first element of the first
row times the first element of the first column. We are going to have negative two times three.
Oh, by the way, we are doing C which is a three by one and A is a one by three. So,
when we get done with this multiplication we are going to have a three by three matrix.
Let me give myself plenty of room here. I hope that is plenty of room. We will adjust
it if need be. Row to column, so negative two times three. Row, because we are going
to across that first row, so as we go across and make the first row of this what is going
to be a three by three matrix... I will squeeze it in up here. I need to get across that entire
row of three elements, so row column, row column, negative two times negative two. And
row column, negative two times five. Now we are going to get into the second row of the
new product, the new matrix. We move on to the second row of that first matrix that we
are multiplying by, that C, so second first column, zero times three. Second row second
column, zero times negative two. Second row third column, zero times five. Getting on
to our third row of what is going to be our new three by three matrix, third row first
column. Third row second column. I just want to be consistent. Then finally one times five.
So our new three by three matrix is going to be the matrix of CA, let's just kind of
divide that up there a little bit, this is going to be negative six, and positive four,
negative ten, zero, zero, zero. The we have our last row of three, negative two, and finally
positive five. That is CA. Now, here is a nice example. We have A times C coming out
to be just a one by one matrix of negative one, and then just turning the order of that
multiplication around, we are used to multiplication being commutative, C times A is not even the
same size matrix much less having all of the same numbers. We ended up with a three by
three matrix. Now, you know, having one row one column is not a super special case but
it is a little bit unique. Let's go ahead and multiply a two by two matrix and a two
by three matrix to finish up our last example of our lesson. BAM!!! And our last example,
we are going to take this two by two matrix and multiply it by this two by three matrix.
Giving us, let's see here, two by two and two by three with these numbers of columns
matching up with these numbers of rows our final answer is going to be in the order of
two by three. Let's call that A and B, matrix AB is going to have the order of two by three.
Ok, so let's get this done. AB, oh...hello, I almost fell over. AB. Ok, everything in
the first getting multiplied by...or in each of the corresponding rows getting matched
up with everything in the corresponding columns of matrix B Again, we are keying in on the
idea this is going to be a two by three matrix. We are going to have well, let's see here,
one, two, three, one two three. I may have to erase these lines as we go just to keep
it neat. But I want that element in the first row first column. That means that we are going
to multiply the first row of the first matrix by the first column of the second matrix.
We have negative two times zero plus three times two. Now I want the second row in the
first column. Excuse me, I said that all wrong. I want the second element in the first row...second
column. First row second column, negative two times negative one plus three times four.
Now, the third row in the... :) I know, it is been a little while getting into this video.
Row one third column, row one third column, negative two time three and we have got three
times negative five. Now moving on to the second row first column, second row first
column, one times zero is going to be zero but I am writing this out. So one times zero
and then plus negative five times two. Then we have got second row second column, so second
row second column one times negative one plus negative five times times four. Now finally
we want this element in the second row and third column. I am getting it straight. We
have second row third column, one times three plus negative five times negative five. I
did get that squeezed in there. So, now we have negative two times zero is zero plus
six and we have six. Here we have two plus twelve is going to be fourteen. We have negative
six and negative fifteen, that is going to come out to be negative twenty-one. We have
zero and negative ten. We have negative one and negative twenty which is going to be negative
twenty-one. Then three, negative five and negative five is twenty-five, so three and
twenty five is going to be twenty-eight and there is our new two by three matrix. The
product of doing matrix A times matrix B. Let me check my notes. They say that negative,
let's see here, six, fourteen, negative twenty-one...whatever. Yes, it checks out. Now properties of matrix
multiplication. If A, B, and C are matrices and c is a scalar, then the following are
true if the order of the matrices allow you to do this. We have the associative property
dealing with the multiplication of matrices. You can regroup the parenthesis, but again
you cannot change the order of those matrices and expect to get the same answer. That will
happen in a special case when you are checking for inverses in an upcoming lesson, but multiplication
is not commutative with matrices. We can do the distributive property with matrices and
this idea of the scalar multiple. If we have matrix A times matrix B and there is a scalar
that is going to be applied to that product, you can apply that scalar to let's say matrix
A first and then multiply that to matrix B and those will be equivalent answers. This
is the end of my lesson. I am Mr. Tarrou. BAM!!! Go Do Your Homework:D