Lesson Video: Introduction to Matrices | Nagwa Lesson Video: Introduction to Matrices | Nagwa

# Lesson Video: Introduction to Matrices Mathematics

In this video, we will learn how to identify matrices and determine the order of a matrix and the position of each of its elements.

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### Video Transcript

In this lesson, weβll 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 theyβre 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, letβs 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 πth row and πth column, which is a scalar quantity, such as a constant or single-valued expression, is π subscript ππ. 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, itβs 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 weβre not going to look at these into much detail. They are the identity matrix and the null matrix. In the identity matrix πΌ, 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 weβve considered the basics of matrices. Letβs 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, letβs 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 weβre 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 thereβs 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.

Weβll 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΄, letβs 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 weβre given a lot of information about each of the elements, but letβs recall how we fill the matrices in. An π-by-π matrix will look something like this, where π subscript ππ is the element that appears in the πth row and πth column. And so letβs 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 weβre given information about π sub one two based on information about π sub three one. And we havenβt filled that in yet, so letβs 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, thatβs here. Then weβre told that π sub two two is a half π sub one one. Thatβs the element in the second row and the second column. So thatβs 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 weβre told π sub three one is equal to eight. Remember, thatβs 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. Thatβs the element in the first row and the second column. Itβs here. Itβs 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. Thereβs one further piece of information, and thatβs 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. Thatβs 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.

Weβre 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 πΆ 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 ππ sits in the πth row and the πth column of that matrix. And so weβre 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 πΆ. 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. Thatβs in matrix π΅, and itβs in the second row and the first column. The second row is here, and the first column is here. The element thatβs 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. Itβs in the second row, which is here, and the first column. Now, of course, itβs 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, weβve been working with the matrix defined as π΄, whose element is π sub ππ. When weβre 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 ππ. And for matrix πΆ, the elements are π sub ππ.

In our next example, weβre going to go back to forming a matrix. But this time, weβre 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¦.

Letβs first consider this information about the order of the matrix. We notice that itβs three by three, and so itβs a square matrix. And of course, this also means that it must have three rows and three columns. Thereβs 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 ππ. Now thatβs the element in the πth row and the πth 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, weβre going to let π₯ be equal to one and π¦ be equal to one. And then weβre 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; itβs nine. Weβre now going to go to the second element in our first row. And this time, weβre 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. Letβs 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 weβve completed the first row of our matrix. Weβre 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 thatβs 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. Weβre 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 weβre going to look at is that we can multiply them by a scalar. Letβs take the matrix π΄ as shown. Weβre 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 ππ sub one one, ππ sub one two, and so on. Letβs 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. Weβre 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.

Letβs 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 weβve completed the matrix that represents the new prices of the drinks. Itβs three, 11, four, 17, seven, 18.

Weβre now going to recap the key points from this lesson. In this video, weβve 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 πth row and πth column is π sub ππ. We saw that if π is equal to π, we say that the matrix is square. Otherwise, itβs 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. Letβs say we want to multiply the matrix π΄ by the scalar π. We do this by multiplying each individual element by π.