Video Transcript
Consider the matrices 𝐴 is the
matrix zero, one, 𝐵 is the matrix negative four, one, negative six, six, 𝐶 is the
matrix five, three. Find 𝐴𝐶𝐵 and 𝐵𝐴𝐶 if
possible.
So, in this problem, what we’re
actually considering is matrix multiplication. So first of all, we need to
remember something about matrix multiplication. And that is if we’re multiplying
matrices together, then the number of columns in the first matrix must be equal to
the number of rows in the second matrix. Well, if we take a look at our
three matrices, matrix 𝐴 is a two-by-one matrix. That means we’ve got two rows and
one column. Matrix 𝐵 is a two-by-two
matrix. And matrix 𝐶 is a one-by-two
matrix.
First of all, we’re going to begin
by looking at 𝐴𝐶𝐵 and seeing whether actually this multiplication is
possible. Well if we start by multiplying 𝐴
and 𝐶, then we see that this would work because we can see that the number of
columns in matrix 𝐴 is one and the number of rows in matrix 𝐶 is also one. And we can also see that if we did
𝐶𝐵 first, this would also work. And that’s because the number of
columns in matrix 𝐶 is two and the number of rows in matrix 𝐵 is also two. So we can see that we could’ve
multiplied either of these combinations first. What we’re going to start with is
matrix 𝐴 multiplied by matrix 𝐶.
So if we consider 𝐴𝐶, this is
going to be the matrix zero, one multiplied by the matrix five, three. And what we know about multiplying
a two-by-one matrix by a one-by-two matrix is that our result matrix is going to be
a two-by-two matrix because we’re gonna have the number of rows from the first
matrix and the number of columns from the second matrix. Now, to form the first element of
our answer matrix, what we do is multiply the corresponding elements from the first
row and first column of our first and second matrices, respectively. So we’re gonna have zero multiplied
by five. So then for our next element, we’re
going to have zero multiplied by three. And that’s because it’s still our
first row, but now we’re looking at the second column in our second matrix.
So then, if we take a look at the
first element from the second row of the first matrix and multiply it by the first
element of the first column of the second matrix, then we’re gonna have one
multiplied by five. And then our final element is found
by multiplying one and three. And then this gives us an answer
matrix which is a two-by-two matrix zero, zero, five, three, and this is 𝐴𝐶.
Okay, so now what we need to do to
find 𝐴𝐶𝐵 is multiply this by 𝐵. So what we’re gonna have is the
matrix zero, zero, five, three multiplied by the matrix negative four, one, negative
six, six. And as we’re multiplying a
two-by-two matrix by a two-by-two matrix, then the answer matrix will also be a
two-by-two matrix. So in order to work out the
different elements for our answer matrix, what we’ll do is multiply the elements
from the first row in the first matrix by the corresponding elements in the first
column in the second matrix, and then we will add them together. So we have zero multiplied by
negative four plus zero multiplied by negative six.
And then we repeat this for the
next element but this time multiplying the elements from the first row in the first
matrix by the second column in the second matrix and then adding them together. So we have zero multiplied by one
plus zero multiplied by six. And then following this pattern on,
we’ll have five multiplied by negative four plus three multiplied by negative six
for the next element and then finally five multiplied by one plus three multiplied
by six for the final element. And then if we calculate each of
these elements, we’ll have the answer matrix zero, zero, negative 38, 23. And this is the value for
𝐴𝐶𝐵.
Okay, now, we move on to the next
part of the problem, and we’ll try and find 𝐵𝐴𝐶. Now in 𝐵𝐴𝐶, if we check again,
if we can do 𝐵𝐴 to begin with, well, we’d be able to do 𝐵𝐴 because the number of
columns in the first matrix 𝐵 is two and the number of rows in the second matrix 𝐴
is also two. So we’ll be able to multiply them
together. Similarly, if we wanted to multiply
𝐴 and 𝐶 first, we’d also be able to do this because the number of columns in the
first matrix 𝐴 is one and the number of rows in the second matrix 𝐶 is also
one.
At this point, it’s worth showing
an example where we wouldn’t be able to find the multiplication. And that would be if we had 𝐴𝐵
because if we take a look at 𝐴𝐵, 𝐴 has one column, whereas matrix 𝐵 has two
rows. So therefore, these are different,
so we would not be able to do the multiplication 𝐴𝐵. Okay, great. Now we’ve seen what we can and
cannot multiply when we’re looking at 𝐵𝐴𝐶. What we’re going to do it is look
at 𝐵𝐴 first.
Now, when we multiply matrix 𝐵 by
matrix 𝐴, we’re gonna have matrix negative four, one, negative six, six multiplied
by matrix zero, one. And this time, because we’re gonna
have a two-by-two matrix multiplied by a two-by-one matrix, our answer matrix is
also gonna be two by one because it’s gonna be the number of rows in the first
matrix by the number of columns in the second matrix.
So to find our values, what we’re
going to do is again multiply our corresponding values from the first row in the
first matrix by the first column in the second matrix and then move on to the second
row in the first matrix by the first row in the second matrix, well the only row in
the second matrix. So we’re gonna have negative four
multiplied by zero plus one multiplied by one and then negative six multiplied by
zero plus six multiplied by one, which is going to give us the answer matrix for
𝐵𝐴 of one, six. Okay, great. Now the final step is now to
multiply 𝐵𝐴 by matrix 𝐶 to give us 𝐵𝐴𝐶.
Well, to do this, what we’re going
to do is multiply matrix one, six by the matrix five, three. And what we have here is a
two-by-one matrix multiplied by a one-by-two matrix. So therefore, the answer matrix is
in fact going to be two by two because it’s the number of rows from the first matrix
by the number of columns in the second matrix.
So then when we carry out the
multiplication, what we’re gonna get is one multiplied by five for our first
element, then one multiplied by three, six multiplied by five, and six multiplied by
three, which is gonna give us the answer matrix five, three, 30, 18. So therefore, we can say that both
𝐴𝐶𝐵 and 𝐵𝐴𝐶 are possible and 𝐴𝐶𝐵 is the matrix zero, zero, negative 38, 23
and 𝐵𝐴𝐶 is the matrix five, three, 30, 18.
