### Video Transcript

In this lesson, weβll learn how to
identify matrices and determine the order of a matrix and the position of its
elements. A matrix is an array of
numbers. And theyβre really useful as they
allow us to represent and work with groups of numbers as if they are one entity. We use them, for instance, to solve
simultaneous equations and represent transformations. We arrange scalars, which we call
elements, in a matrix, plural matrices, in rows and columns. When we describe a matrix, letβs
call that π΄, as an π-by-π matrix, this has π rows and π columns. And we can represent this matrix as
shown.

We can now say that the element
that sits in the πth row and πth column, which is a scalar quantity, such as a
constant or single-valued expression, is π subscript ππ. For example, the element in the
third row and the fifth column is π subscript three five. Now it also follows that if π is
equal to π, in other words, the number of rows and columns are equal, the matrix is
square. Otherwise, itβs rectangular. Similarly, if either π or π is
equal to one, then we say we have a vector. And we use these to define points
in space.

Now, in fact, there are two special
matrices that we should be aware of, although weβre not going to look at these into
much detail. They are the identity matrix and
the null matrix. In the identity matrix πΌ, all of
the elements are equal to zero except those in the main diagonal, as shown, which
have a value of one. And in the null matrix π, each of
the elements are equal to zero. And so now weβve considered the
basics of matrices. Letβs have a look at a few
questions that introduce us to these concepts.

How many elements are there in a
matrix of order nine by seven?

Remember, we arrange scalars, which
we call elements, in a matrix in rows and columns. When we describe a matrix, letβs
call this general matrix π΄, as an π-by-π matrix, it has π rows and π
columns. And this matrix would look a little
something like this. The order of this matrix is π by
π. Now weβre given a matrix of order
nine by seven. And so this must mean that our
matrix has nine rows and seven columns. So how many elements must it
have?

Well, if we have nine rows and each
of those rows has seven columns and thereβs an element in each of these, then we can
find the number of elements by multiplying nine by seven. Nine times seven is equal to
63. And so, in a matrix of order nine
by seven, there must be a total of 63 elements. And of course, it follows that we
can generalize this and say that a matrix π-by-π order must have π times π
elements.

Weβll now consider how we can
complete a matrix given definitions of each of its elements.

Given that π΄ is a matrix of order
three by two, where π sub one one is equal to zero, π sub one two is equal to π
sub three one minus three, π sub two one is equal to four, π sub two two is equal
to half π sub one one, π sub three one is equal to eight, and π sub three two is
equal to a quarter π sub two one, determine π΄.

In order to write out the matrix
π΄, letβs begin by identifying how many rows and columns it has. We know that a matrix of order π
by π has π rows and π columns. This matrix π΄ has order three by
two. And so we can see that it must have
a total of three rows and two columns. So there will be three times two
which is equal to six elements. Now weβre given a lot of
information about each of the elements, but letβs recall how we fill the matrices
in. An π-by-π matrix will look
something like this, where π subscript ππ is the element that appears in the πth
row and πth column. And so letβs consider π sub one
one. This element will appear in the
first row and the first column.

π sub one one is equal to
zero. So we put a zero in the top
left-hand corner of our matrix. Now weβre given information about
π sub one two based on information about π sub three one. And we havenβt filled that in yet,
so letβs move on to the next bit of information. π sub two one is equal to
four. This tells us that the element in
the second row and first column is equal to four. Well, thatβs here. Then weβre told that π sub two two
is a half π sub one one. Thatβs the element in the second
row and the second column. So thatβs here. We know that π sub one one is
equal to zero and a half of zero is still zero. So we add zero in this place.

Then weβre told π sub three one is
equal to eight. Remember, thatβs the element in the
third row and the first column. So we have an eight here. We can now go back to the
information about π sub one two. Thatβs the element in the first row
and the second column. Itβs here. Itβs π sub three one which is
equal to eight minus three. And since eight minus three is
five, we put a five in the top-right corner of our matrix. Thereβs one further piece of
information, and thatβs π sub three two is a quarter π sub two one. Well, π sub three two is the
element in the third row and the second column. Thatβs here.

We saw that π sub two one is equal
to four. So we need to find a quarter of
four, which is equal to one. When we read matrices, we read
along the columns from left to right. And so our matrix π΄ is zero, five,
four, zero, eight, one.

Weβre now going to further develop
this skill by looking at how we can find entries in a given matrix.

Given that the matrix π΄ is equal
to negative two, four, negative seven, negative one, nine, nine, the matrix π΅ is
equal to negative seven, three, two, negative six, negative four, negative eight,
seven, three, zero, and πΆ is equal to two, negative five, negative four, find π
sub two three, π sub two one, and π sub two one.

We begin by recalling how we define
the elements in a matrix. The element defined by π
subscript ππ sits in the πth row and the πth column of that matrix. And so weβre looking to find π sub
two three, that will be an element from our first matrix π΄, π sub two one, that
will be an element from the matrix π΅, and π sub two one, which will be an element
of the matrix πΆ. If we compare π sub two three with
the general definition, we see that π sub two three must be the element in the
second row and the third column. The second row of matrix π΄ is here
and the third column is here. The element that sits in the
intersection of these is nine. And so π sub two three must be
equal to nine.

Next, we have π sub two one. Thatβs in matrix π΅, and itβs in
the second row and the first column. The second row is here, and the
first column is here. The element thatβs in the
intersection of these is negative six. And so we see that π sub two one
is equal to negative six. We have the same situation with π
sub two one. Itβs in the second row, which is
here, and the first column. Now, of course, itβs only one
column. The element in the intersection of
these is negative five. And so π sub two one must indeed
be equal to negative five. And so π sub two three is nine, π
sub two one is negative six, and π sub two one is equal to negative five.

Now notice that up until this
point, weβve been working with the matrix defined as π΄, whose element is π sub
ππ. When weβre working with multiple
matrices, we can use the lowercase letter to describe each of the elements. So for matrix π΅, we described its
elements by π sub ππ. And for matrix πΆ, the elements are
π sub ππ.

In our next example, weβre going to
go back to forming a matrix. But this time, weβre going to use
an equation to find each of its elements.

Find the matrix π΄ which is equal
to elements π sub π₯π¦ with an order of three by three whose elements are given by
the formula π sub π₯π¦ equals five π₯ plus four π¦.

Letβs first consider this
information about the order of the matrix. We notice that itβs three by three,
and so itβs a square matrix. And of course, this also means
that it must have three rows and three columns. Thereβs going to be a total of
three times three, which is nine elements to find in this matrix. And so next, we recall that we
define the elements in a matrix as π sub ππ. Now thatβs the element in the πth
row and the πth column. And so the element in the first
row and first column is π sub one one. The element in the first row and
second column is π sub one two, and the third element in this row is π sub one
three. In the second row and first column,
we have π sub two one. And then we can complete the rest
of the matrix as shown.

Now notice that our matrix is
defined by elements π sub π₯π¦. And so to find the element π sub
one one, weβre going to let π₯ be equal to one and π¦ be equal to one. And then weβre going to use this
formula here, π sub π₯π¦ is five π₯ plus four π¦. We simply substitute our values of
π₯ and π¦ into this formula, and we get π sub one one is five times one plus four
times one, which is nine. And so we have the first element in
our matrix; itβs nine. Weβre now going to go to the second
element in our first row. And this time, weβre going to let
π₯ be equal to one and π¦ be equal to two.

This time, our formula is five
times one plus four times two which is equal to 13. And so 13 is the second element in
the first row of our matrix. Letβs now consider this
element. This time, we let π₯ be equal to
one and π¦ be equal to three. And so we get π sub one three must
be equal to five times one plus four times three, which is 17. And so weβve completed the first
row of our matrix. Weβre now going to move on to the
second row of our matrix. Notice that for each element here,
π₯ is always equal to two. And so π sub two one is found by
letting π₯ be equal to two and π¦ be equal to one.

We get five times two plus four
times one, which is 14. Then π sub two two is five times
two plus four times two which is 18. And π sub two three is five times
two plus four times three, which is 22. And so we complete the second row
of our matrix. All thatβs left is to find the
values in the third row. This time, π₯ is always equal to
three. And so the first element is five
times three plus four times one, which is 19. The second element is five times
three plus four times two, which is 23. And our third element is five times
three plus four times three, which is 27. And so we complete the third row in
our matrix. Weβre therefore able to say that
given the definition for matrix π΄, we get nine, 13, 17, 14, 18, 22, 19, 23, 27.

Now there are several operations
that we can apply to matrices, and it is outside the scope of this video to look at
them all. However, one that weβre going to
look at is that we can multiply them by a scalar. Letβs take the matrix π΄ as
shown. Weβre going to multiply this by the
scalar π. And to do this, we can simply
multiply each individual element by π. So π times π΄ is equal to ππ sub
one one, ππ sub one two, and so on. Letβs have a look at one example of
the application of this.

The table below represents the
prices of some drinks in a cafe. The cafe owner changes the prices
of the drinks so that each drink is now two times its original price. Determine the matrix that
represents the new prices of the drinks.

A matrix is simply an array of
numbers. And so, if we define π΄ to be the
matrix that represents the old prices of the drinks, we can simply transfer each of
the numbers from our table in as shown. The order of this matrix is three
by two, since it has three rows and two columns. And so matrix π΄ is 1.5, 5.5, two,
8.5, 3.5, nine. Weβre told that the prices of the
drinks changes so that the price of each drink is now two times its original
price. And so, we want to double. We want to multiply each element in
the matrix by two. We can represent this as two π΄,
since we know that when we multiply a matrix by a scalar, we multiply each of the
individual elements. So the matrix two π΄ represents the
new prices of the drinks. And its elements are two times 1.5,
two times 5.5, two times two, two times 8.5, two times 3.5, and two times nine.

Letβs complete this element by
element. Two times 1.5 is three. So three is the element in the
first row and first column. Then two times 5.5 is 11, two times
two is four, and two times 8.5 is 17. Two times 3.5 is seven, and two
times nine is 18, meaning the element in our third row and second column is 18. And so weβve completed the matrix
that represents the new prices of the drinks. Itβs three, 11, four, 17, seven,
18.

Weβre now going to recap the key
points from this lesson. In this video, weβve learned that a
matrix is an array of numbers. And elements within the matrix are
arranged in rows and columns. Given a matrix π΄ which has order
π by π, this means it has π rows and π columns. Then the matrix π΄ will look a
little something like this, where we say that the element in the πth row and πth
column is π sub ππ. We saw that if π is equal to π,
we say that the matrix is square. Otherwise, itβs a rectangular
matrix. And if either π or π is equal to
one, then we call this a vector. And we use these to define points
in space.

We saw that there are two special
matrices called the identity matrix and the null matrix. The identity matrix is a square
matrix where all elements are zero except those on the main diagonal, in other
words, where π is equal to π, where those have a value of one. And the null matrix π has a value
of zero for all its elements. Finally, we saw that we can
multiply a matrix by a scalar. Letβs say we want to multiply the
matrix π΄ by the scalar π. We do this by multiplying each
individual element by π.