WEBVTT
00:00:00.480 --> 00:00:06.840
In this video, we will learn how to use matrix multiplication to determine the square and cube of a square matrix.
00:00:07.520 --> 00:00:16.200
Whilst the methods we use can be extended to higher powers, in this video, we will only deal with squaring and cubing.
00:00:16.640 --> 00:00:22.880
It is important to note that we can raise any π-by-π square matrix to any power.
00:00:23.320 --> 00:00:29.800
In this video, we will only deal with two-by-two and three-by-three matrices.
00:00:30.400 --> 00:00:32.560
Letβs begin with some key definitions.
00:00:33.040 --> 00:00:44.400
For a square matrix π΄ and positive integr π, the πth power of π΄ is defined by multiplying this matrix by itself repeatedly.
00:00:44.880 --> 00:00:55.280
That is, π΄ to the πth power is equal to π΄ multiplied by π΄ multiplied by π΄, and so on, multiplied by π΄, where there are π instances of the matrix π΄.
00:00:55.800 --> 00:01:00.960
In this video, we will calculate the matrix π΄ squared by multiplying matrix π΄ by matrix π΄.
00:01:01.440 --> 00:01:07.440
We will also calculate π΄ cubed by multiplying matrix π΄ by itself and itself again.
00:01:08.000 --> 00:01:11.920
This is equivalent to the matrix π΄ multiplied by the matrix π΄ squared.
00:01:12.480 --> 00:01:16.840
Alternatively, we can multiply the matrix π΄ squared by the matrix π΄.
00:01:17.280 --> 00:01:25.080
It is important to note that taking the power of a matrix is only well defined if the matrix is square.
00:01:25.360 --> 00:01:33.280
If matrix π΄ has order π by π, then this order will be common to π΄ squared, π΄ cubed, and so on.
00:01:33.640 --> 00:01:37.640
Squaring and cubing a matrix does not alter its order.
00:01:38.040 --> 00:01:45.640
As already mentioned, in this video, we will focus on two-by-two and three-by-three square matrices.
00:01:46.120 --> 00:01:51.160
We will now look at an example where we need to square a two-by-two matrix.
00:01:51.920 --> 00:01:58.160
Given that matrix π΄ equals negative six, one, negative five, five, find π΄ squared.
00:01:58.680 --> 00:02:07.280
We recall that when dealing with powers of matrices, the matrix π΄ squared will only be defined if the matrix π΄ is square.
00:02:07.880 --> 00:02:11.520
In this case, π΄ is a two-by-two matrix.
00:02:11.920 --> 00:02:15.960
To calculate π΄ squared, we need to multiply the matrix π΄ by itself.
00:02:16.440 --> 00:02:23.400
This is equal to negative six, one, negative five, five multiplied by negative six, one, negative five, five.
00:02:23.800 --> 00:02:31.800
When multiplying matrices, we begin by multiplying the elements in the first row of the first matrix by the first column in the second matrix.
00:02:32.400 --> 00:02:39.800
The top-left element or component of our matrix will be equal to negative six multiplied by negative six plus one multiplied by negative five.
00:02:40.240 --> 00:02:45.800
Next, we multiply the first row of the first matrix by the second column of the second matrix.
00:02:46.360 --> 00:02:50.480
This gives us negative six multiplied by one plus one multiplied by five.
00:02:50.880 --> 00:02:55.400
We can then repeat this process for the second row of the first matrix.
00:02:55.960 --> 00:03:04.760
This gives us negative five multiplied by negative six plus five multiplied by negative five and negative five multiplied by one plus five multiplied by five.
00:03:05.240 --> 00:03:12.800
Simplifying each of our elements gives us the matrix 31, negative one, five, 20.
00:03:13.480 --> 00:03:22.640
If matrix π΄ is equal to negative six, negative one, negative five, five, then π΄ squared is equal to 31, negative one, five, 20.
00:03:23.320 --> 00:03:25.720
We notice that the order of the matrix stays the same.
00:03:26.040 --> 00:03:29.680
And this is true when we raise any square matrix to any power.
00:03:30.160 --> 00:03:36.520
We will now consider what happens when we need to square and cube a three-by-three matrix.
00:03:37.360 --> 00:03:45.400
Consider the matrix π΄ which is equal to one, one, two, one, zero, one, two, one, zero.
00:03:45.800 --> 00:03:47.280
There are two parts to this question.
00:03:47.800 --> 00:03:52.080
Firstly, find π΄ squared, and secondly, find π΄ cubed.
00:03:52.960 --> 00:03:57.240
We recall that in order to square any matrix, we multiply it by itself.
00:03:57.600 --> 00:04:04.000
Also for the matrix π΄ squared to be well defined, matrix π΄ must be a square matrix.
00:04:04.560 --> 00:04:08.320
In this case, it is a three-by-three matrix.
00:04:08.640 --> 00:04:21.000
This means that π΄ squared is equal to one, one, two, one, zero, one, two, one, zero multiplied by one, one, two, one, zero, one, two, one, zero.
00:04:21.440 --> 00:04:29.400
When multiplying two matrices, we begin by multiplying the elements in the first row of the first matrix by the first column of the second matrix.
00:04:30.040 --> 00:04:36.120
The element in the top-left corner will be equal to one times one plus one times one plus two times two.
00:04:36.680 --> 00:04:37.800
This is equal to six.
00:04:38.440 --> 00:04:45.320
Next, we multiply the elements of the first row of the first matrix by the second column of the second matrix.
00:04:45.760 --> 00:04:53.680
This gives us one multiplied by one plus one multiplied by zero plus two multiplied by one, which is equal to three.
00:04:54.320 --> 00:05:08.640
Moving on to the third column of the second matrix, we also get an answer of three, as one multiplied by two plus one multiplied by one plus two multiplied by zero equals three.
00:05:09.120 --> 00:05:13.120
We then repeat this process for the second row of the first matrix.
00:05:13.520 --> 00:05:16.640
This gives us answers of three, two, and two.
00:05:17.320 --> 00:05:26.880
Finally, multiplying the third row of the first matrix by each of the columns in the second matrix gives us values three, two, and five.
00:05:27.400 --> 00:05:37.160
π΄ squared is the three-by-three matrix six, three, three, three, two, two, three, two, five.
00:05:37.760 --> 00:05:41.880
The second part of our question asks us to find the matrix π΄ cubed.
00:05:42.520 --> 00:05:50.600
We can do this by multiplying matrix π΄ by the matrix π΄ squared or by multiplying the matrix π΄ squared by the matrix π΄.
00:05:51.120 --> 00:05:53.400
In this question, we will use the first method.
00:05:53.760 --> 00:06:05.840
π΄ cubed is equal to one, one, two, one, zero, one, two, one, zero multiplied by six, three, three, three, two, two, three, two, five.
00:06:06.360 --> 00:06:10.640
We use the same method as the first part of the question.
00:06:11.080 --> 00:06:17.640
We begin by multiplying the elements of the first row of the first matrix by the first column of the second matrix.
00:06:18.200 --> 00:06:24.640
One multiplied by six plus one multiplied by three plus two multiplied by three is equal to 15.
00:06:25.120 --> 00:06:31.160
Multiplying the first row of the first matrix by the second column of the second matrix gives us nine.
00:06:31.480 --> 00:06:35.960
And multiplying by the third column of the second matrix gives us 15.
00:06:36.560 --> 00:06:41.640
The second row of the matrix π΄ cubed has elements nine, five, and eight.
00:06:42.200 --> 00:06:47.440
Finally, the third row is equal to 15, eight, and eight.
00:06:47.920 --> 00:06:56.240
The matrix π΄ cubed is equal to 15, nine, 15, nine, five, eight, 15, eight, eight.
00:06:56.800 --> 00:07:11.080
If weβre given any square matrix π΄, we can calculate π΄ squared by multiplying the matrix by itself and π΄ cubed by multiplying matrix π΄ by the matrix π΄ squared.
00:07:11.600 --> 00:07:16.600
In our next question, we will simplify an expression by using squared and cubed matrices.
00:07:17.240 --> 00:07:24.640
Given the matrix π΄, which is equal to four, zero, negative three, seven, calculate π΄ cubed minus three π΄ squared.
00:07:25.400 --> 00:07:33.680
We recall that when given any square matrix π΄, we can calculate the matrix π΄ squared by multiplying π΄ by itself.
00:07:34.280 --> 00:07:41.840
In this question, π΄ squared is equal to four, zero, negative three, seven multiplied by four, zero, negative three, seven.
00:07:42.440 --> 00:07:50.520
When multiplying matrices, we begin by multiplying the elements in the first row of the first matrix by the elements in the first column of the second matrix.
00:07:51.000 --> 00:07:56.080
Four multiplied by four plus zero multiplied by negative three is equal to 16.
00:07:56.520 --> 00:08:01.480
We then repeat this for the second column of the second matrix.
00:08:01.960 --> 00:08:06.160
Four multiplied by zero plus zero multiplied by seven is equal to zero.
00:08:06.800 --> 00:08:12.520
Next, we multiply the second row of the first matrix by each of the columns in the second matrix.
00:08:13.080 --> 00:08:16.480
This gives us negative 33 and 49.
00:08:17.280 --> 00:08:26.680
π΄ squared is equal to the two-by-two matrix 16, zero, negative 33, 49.
00:08:27.320 --> 00:08:30.640
In our expression, we want three π΄ squared.
00:08:30.920 --> 00:08:37.080
This means we need to multiply the matrix π΄ squared by the scalar or constant three.
00:08:37.640 --> 00:08:46.680
We multiply each of the elements by three, giving us 48, zero, negative 99, 147.
00:08:47.280 --> 00:08:53.000
Next, we need to calculate π΄ cubed, and we know this is equal to π΄ multiplied by π΄ squared.
00:08:53.560 --> 00:09:03.760
π΄ cubed is therefore equal to four, zero, negative three, seven multiplied by 16, zero, negative 33, 49.
00:09:04.360 --> 00:09:17.640
We multiply these two matrices in the same way as we calculated π΄ squared, giving us π΄ cubed is equal to 64, zero, negative 279, 343.
00:09:18.480 --> 00:09:22.560
We now have a matrix for π΄ cubed and also for three π΄ squared.
00:09:22.880 --> 00:09:24.960
We are asked in the question to subtract these.
00:09:25.440 --> 00:09:38.680
We have 64, zero, negative 279, 343 minus 48, zero, negative 99, 147.
00:09:39.240 --> 00:09:42.920
When subtracting matrices, we subtract the individual components or elements.
00:09:43.400 --> 00:09:47.560
This means that we begin by subtracting 48 from 64.
00:09:48.200 --> 00:09:49.560
This is equal to 16.
00:09:50.080 --> 00:09:53.720
Subtracting the top-right elements gives us zero.
00:09:54.160 --> 00:10:02.760
Negative 279 minus negative 99 is the same as adding 99 to negative 279.
00:10:03.320 --> 00:10:05.360
This is equal to negative 180.
00:10:05.920 --> 00:10:13.120
Finally, 343 minus 147 is equal to 196.
00:10:13.760 --> 00:10:26.720
If the matrix π΄ is equal to four, zero, negative three, seven, then π΄ cubed minus three π΄ squared is equal to 16, zero, negative 180, 196.
00:10:27.680 --> 00:10:33.320
In our final question, we will solve a system of linear equations by applying operations on matrices.
00:10:34.120 --> 00:10:54.040
Given that the matrix π is equal to five, six, negative five, negative four, find the values of π₯ and π¦ that satisfy π squared plus π₯π plus π¦πΌ is equal to π, where π is the zero matrix of order two by two and πΌ is the unit matrix of order two by two.
00:10:54.960 --> 00:10:59.120
We recall that any zero matrix has all elements equal to zero.
00:10:59.520 --> 00:11:03.920
Therefore, the matrix π is equal to zero, zero, zero, zero.
00:11:04.360 --> 00:11:10.560
The unit or identity matrix has ones on its main or leading diagonal and zeros elsewhere.
00:11:10.880 --> 00:11:14.800
Therefore, πΌ is equal to one, zero, zero, one.
00:11:15.160 --> 00:11:25.320
When multiplying any matrix by a constant, in this question π₯ and π¦, we simply multiply each of the elements in the matrix by the constant.
00:11:25.800 --> 00:11:30.680
This means that the matrix π¦πΌ is equal to π¦, zero, zero, π¦.
00:11:31.440 --> 00:11:41.440
As π is equal to five, six, negative five, negative four, then π₯π is equal to five π₯, six π₯, negative five π₯, negative four π₯.
00:11:42.120 --> 00:11:44.840
Finally, we need to calculate the matrix π squared.
00:11:45.200 --> 00:11:48.200
This is equal to matrix π multiplied by itself.
00:11:48.680 --> 00:11:55.600
We need to multiply five, six, negative five, negative four by five, six, negative five, negative four.
00:11:56.080 --> 00:12:04.800
When multiplying matrices, we multiply each of the rows in the first matrix by each of the columns in the second matrix.
00:12:05.120 --> 00:12:10.400
Five multiplied by five plus six multiplied by negative five is equal to negative five.
00:12:10.880 --> 00:12:17.840
Multiplying the elements of the first row in the first matrix by the second column in the second matrix gives us six.
00:12:18.280 --> 00:12:26.280
Repeating this for the second row of the first matrix, we get the elements negative five and negative 14.
00:12:26.960 --> 00:12:44.320
Substituting the four matrices into our equation, we have negative five, six, negative five, negative 14 plus five π₯, six π₯, negative five π₯, negative four π₯ plus π¦, zero, zero, π¦ is equal to zero, zero, zero, zero.
00:12:45.160 --> 00:12:49.920
We can now set up four equations comparing the corresponding components or elements.
00:12:50.560 --> 00:12:57.240
Comparing the top-left elements, we have negative five plus five π₯ plus π¦ is equal to zero.
00:12:57.760 --> 00:12:59.560
We will call this equation one.
00:13:00.000 --> 00:13:06.800
The top-right elements give us six plus six π₯ plus zero is equal to zero.
00:13:07.560 --> 00:13:13.080
As there was only one unknown in this equation, we can solve it.
00:13:13.560 --> 00:13:17.880
Subtracting six from both sides gives us six π₯ is equal to negative six.
00:13:18.360 --> 00:13:23.720
Dividing both sides of this equation by six gives us π₯ is equal to negative one.
00:13:24.480 --> 00:13:29.360
We can then substitute this value back into equation one to calculate the value of π¦.
00:13:30.200 --> 00:13:35.040
Simplifying the left-hand side, we have negative 10 plus π¦ is equal to zero.
00:13:35.640 --> 00:13:37.200
This means that π¦ is equal to 10.
00:13:37.640 --> 00:13:42.800
The values of π₯ and π¦ are negative one and 10, respectively.
00:13:43.400 --> 00:13:49.920
We do need to check, however, that these values hold for the bottom row of our matrices.
00:13:50.520 --> 00:14:01.160
Here, we have the two equations: negative five plus negative five π₯ plus zero is equal to zero and negative 14 plus negative four π₯ plus π¦ is equal to zero.
00:14:01.720 --> 00:14:07.440
Adding five π₯ to both sides of the first equation gives us five π₯ is equal to negative five.
00:14:07.760 --> 00:14:12.960
Dividing both sides by five, we see that π₯ once again is equal to negative one.
00:14:13.560 --> 00:14:21.600
In the second equation, substituting π₯ is negative one gives us negative 14 plus four plus π¦ is equal to zero.
00:14:22.160 --> 00:14:24.840
Once again, this confirms that π¦ is equal to 10.
00:14:25.360 --> 00:14:40.320
If the matrix π equals five, six, negative five, negative four, then the values of π₯ and π¦ that satisfy the equation π squared plus π₯π plus π¦πΌ equals π are π₯ is equal to negative one and π¦ is equal to 10.
00:14:41.040 --> 00:14:44.600
We will now summarize the key points from this video.
00:14:45.280 --> 00:14:58.800
For a square matrix π΄ and positive integer π, π΄ to the πth power is equal to π΄ multiplied by π΄ multiplied by π΄, and so on, multiplied by π΄, where there are π instances of π΄.
00:14:59.320 --> 00:15:04.160
The power of a matrix is only well defined if the matrix is square.
00:15:04.840 --> 00:15:10.920
In this video, we dealt with two-by-two and three-by-three square matrices.
00:15:11.240 --> 00:15:15.600
However, this extends to any π-by-π square matrix.
00:15:16.320 --> 00:15:30.280
In this video, we used the general rule to square and cube matrices, where π΄ squared is equal to π΄ multiplied by π΄ and π΄ cubed is equal to π΄ squared multiplied by π΄ or π΄ multiplied by π΄ squared.
00:15:30.880 --> 00:15:35.360
This method can also be extended when dealing with higher powers.