### Video Transcript

In this video, we will learn how to
identify the conditions for matrix multiplication and evaluate the product of two
matrices if possible. Recall that a matrix is an array,
often consisting of numbers which we call elements, but also a matrix could consist
of symbols or expressions. We often use capital letters to
represent matrices. We can describe the size of a
matrix with its dimensions. If a matrix has π rows and π
columns, we say that this is an π-by-π matrix. For example, this is a two-by-two
matrix and this is a three-by-four matrix.

There are two types of
multiplication we can do with matrices, scalar multiplication and the matrix
multiplication. Scalar multiplication involves
multiplying a matrix by a scalar. This just means multiplying a
matrix by a number. For example, with the matrix π΅, we
could find three π΅ by multiplying each component of the matrix π΅ by three. So that is scalar
multiplication. Matrix multiplication is a bit
harder than this as it involves multiplying matrices together. In order to do matrix
multiplication, weβve got to pay attention to the size of the matrices which we want
to multiply together. We canβt just multiply any two
matrices together.

For two matrices π΄ and π΅, to find
the product π΄π΅, if π΄ has dimensions π by π, then matrix π΅ must have dimensions
π by π in order for multiplication to work. In other words, matrix π΄ must have
the same number of columns as the rows of matrix π΅. We can also determine the size of
the resultant matrix π΄π΅. The resultant matrix will have
dimensions π by π. Letβs demonstrate this in an
example.

Consider the one-by-two matrix π΄
and the two-by-one matrix π΅. Their product π΄π΅ must exist
because π΄ has the same number of columns as the number of rows in matrix π΅. We can also say that π΄π΅ has
dimensions one by one. The multiplication of these two
matrices is very similar to the way we find the dot product. We start by doing two multiplied by
seven and then we add three multiplied by one. This gives us 14 add three, which
is 17.

Letβs now have a look at a more
difficult example.

Consider the matrices π΄ and
π΅. Find π΄π΅, if possible.

Letβs first establish whether the
multiplication of these two matrices is possible. Matrix π΄ has three rows and two
columns, and matrix π΅ has two rows and three columns. Seen is the number of columns in
matrix π΄ is the same as the number of rows in matrix π΅, we know that the resultant
matrix exists. Additionally, we can tell the
dimensions of the resultant matrix by seeing that matrix π΄ has three rows and
matrix π΅ has three columns. So the resultant matrix will be a
three-by-three matrix. We find the first element of π΄π΅
by multiplying the top row of matrix π΄ by the left-hand column of matrix π΅. Remember that this is just the same
as finding the dot product of this first row of matrix π΄ with the left-hand column
of matrix π΅.

That is 11 multiplied by negative
eight add negative two multiplied by negative four. To get the top middle element, we
multiply the top row of matrix π΄ with the middle column of matrix π΅. That is 11 multiplied by negative
nine add negative two multiplied by eight. To get the top right element, we
multiply the top row of matrix π΄ with the right-hand column of matrix π΅. That is 11 multiplied by six add
negative two multiplied by nine. We can then find the middle left
component by multiplying the middle row of matrix π΄ with the left-hand column of
matrix π΅.

We find the middle component by
multiplying the middle row of matrix π΄ with the middle column of matrix π΅. And we find the middle right
component by multiplying together the middle row of matrix π΄ with the right-hand
column of matrix π΅. And we follow the same pattern for
the bottom left component, the bottom middle component, and the bottom right
component. We can then simplify each
component. And that gives us our final answer,
which is just as we worked out a three-by-three matrix.

One really important thing to note
is that matrix multiplication is not commutative. This means that π΄π΅ is not equal
to π΅π΄. We can see how this is the case by considering the dimensions of the
matrices in the previous example. If we were to work out the
dimensions of the resultant matrix, we would see that we would end up with a
two-by-two matrix, whereas π΄π΅ gave us a three-by-three matrix. We can use matrix multiplication to
find powers of matrices.

Letβs have a look at an
example.

Given that π΄ equals negative six,
one, negative five, five, find π΄ squared.

Remember that π΄ squared simply
means π΄ multiplied by π΄. So this simply means the matrix π΄
multiplied by the matrix π΄. So because π΄ is a two-by-two
matrix, weβre doing a two-by-two matrix multiplied by a two-by-two matrix. We know this is possible because
the number of columns in matrix π΄ is of course the same as the number of rows in
matrix π΄. And the resultant matrix is going
to be a two-by-two matrix. We find the top-left component of
the resultant matrix by multiplying the top row of the first matrix by the left-hand
column of the second matrix. That is negative six times negative
six add one times negative five.

We then find the top right-hand
component by multiplying the top row of the first matrix by the right-hand column of
the second matrix, that is, negative six times one add one times five. We can then find the bottom left
component by multiplying the bottom row of the first matrix by the left-hand column
of the second matrix. That is negative five times
negative six add five times negative five. And we find the bottom right-hand
element by doing the bottom row of the first matrix multiplied by the right-hand
column of the second matrix. That is negative five times one add
five times five.

The first thing Iβm going to do is
work out each of these multiplications. And weβve got to be really careful
here as we have a lot of negatives. And finally, we can simplify to get
our final answer. π΄ squared equals 31, negative one,
five, 20. One thing to note is that we can
use this process to find higher powers of π΄. For example, we could find π΄ cubed
by multiplying π΄ squared by π΄ on the right. It would look like this.

Letβs see one more example on
matrix multiplication.

Given that π΄ equals negative
three, negative seven, negative one, three, four, one; π΅ equals six, negative four,
three, find π΄π΅ if possible.

Recall that for the multiplication
of matrices π΄ and π΅, if π΄ has dimensions π by π, where π is the number of rows
and π is the number of columns, then π΅ must have dimensions π by π in order for
matrix multiplication to work. In other words, matrix π΄ must have
the same number of columns as the number of rows in matrix π΅. Matrix π΄ has two rows and three
columns, so it has dimensions two by three. Matrix π΅ has three rows and one
column, so it has dimensions three by one. So as the number of columns of
matrix π΄ is the same as the number of rows of matrix π΅, the product exists.

We can also tell the dimensions of
the resultant matrix. For π΄ with dimensions π by π and
π΅ with dimensions π by π, π΄π΅ has dimensions π by π. So for this question, we can tell
that the resulting matrix will have dimensions two by one. Itβs always worth checking this
before beginning a matrix multiplication question to avoid making any mistakes.

Letβs now go ahead and multiply
these matrices together. We find the top component in π΄π΅
by taking the top row of matrix π΄ and multiplying it with the column in matrix
π΅. That is negative three times six
add negative seven times negative four add negative one times three. And we get the bottom row in the
matrix π΄π΅ by taking the bottom row of the matrix π΄ and multiplying it with the
column in matrix π΅. That is three times six add four
times negative four add one times three. We can then multiply the terms
together and then simplify to get the final answer. Just as we expected, the resultant
matrix π΄π΅ has dimensions two by one.

We can use matrix multiplication to
find unknown entries in matrices which are part of a product. As an example, letβs say we know
that the product of the matrices three, two, five, π₯ and one, three equals nine,
negative one. We know that the nine comes from
the multiplication of the top row of the first matrix with the column in the second
matrix. But the negative one comes from the
multiplication of the bottom row of the first matrix with the column in the second
matrix. That is five times one add π₯ times
three. In other words, we know that five
add three π₯ must give us negative one. So three π₯ must be negative
six. Therefore, π₯ equals negative
two.

Letβs see a harder example of
this.

Find the values of π₯ and π¦ given
the following: the matrix one, three, negative two, one multiplied by the matrix
two, zero, π₯, π¦ equals the matrix eight, negative nine, negative two, negative
three.

To solve this for π₯ and π¦, we can
consider how we obtain some of the components and the resultant matrix. Letβs start with the top left
component, which is eight. We know that the eight mustβve been
obtained by the multiplication of the top row of the first matrix with the left-hand
column of the second matrix. That is one times two add three
times π₯ equals eight or two add three π₯ equals eight. We then obtain that three π₯ equals
six by subtracting two from both sides. And we find that π₯ must be equal
to two.

Now if we think about how the
negative nine was obtained, as this is the top right element, this is the
multiplication of the top row of the first matrix with the right-hand column of the
second matrix. That is, one times zero add three
times π¦ must equal negative nine. This simplifies to three π¦ equals
negative nine. Therefore, π¦ must be equal to
negative three. Weβre then able to verify our
answer by checking that these values of π₯ and π¦ work for the bottom two values in
the resultant matrix.

To get the bottom left value of the
resultant matrix negative two, we must do the bottom row of the first matrix
multiplied by the left-hand column of the second matrix. That is negative two times two add
one times π₯, which is two, equals negative two. This gives us negative four add two
equals negative two, which is true. So our value of π₯ is correct. And we can check the bottom
right-hand component of the resultant matrix by multiplying the bottom row of the
first matrix with the right-hand column of the second matrix. That is negative two times zero add
one times negative three equals negative three. That is zero add negative three
equals negative three, which is true. So we know our value for π¦ is also
correct. Itβs always good when possible to
confirm the values that you found for π₯ and π¦.

Letβs now summarize the main points
from this video. There are two types of
multiplication of matrices: scalar multiplication and matrix multiplication. Matrix multiplication is not
commutative. For two matrices π΄ and π΅, π΄
multiplied by π΅ is not equal to π΅ multiplied by π΄. Matrix multiplication is only
possible when the first matrix has the same number of columns as the number of rows
of the second matrix. And we can use the dimensions of
the two matrices weβre multiplying together to find the dimensions of the resultant
matrix. We can use matrix multiplication to
find powers of matrices. And we can use matrix
multiplication to find unknowns in matrix equations by considering how certain
components in the resultant matrix are obtained.