WEBVTT
00:00:00.970 --> 00:00:08.680
In this video, we will learn how to evaluate three-by-three determinants using the cofactors, Laplace expansion, or the SARAS method.
00:00:09.990 --> 00:00:18.950
Calculating the determinant of a square matrix is really useful to gather information about the matrix, such as being able to tell whether or not the matrix is invertible.
00:00:20.180 --> 00:00:25.250
Letβs start with a quick reminder on how to calculate the determinant of two-by-two matrix.
00:00:25.600 --> 00:00:35.510
Then weβll go on to see how we can calculate the determinant of a general square matrix with order π by π using what we know about finding the determinant of a two-by-two matrix.
00:00:36.300 --> 00:00:38.960
Letβs consider this two-by-two matrix.
00:00:39.550 --> 00:00:44.160
The determinant of π΄ is denoted with straight lines either side of π΄.
00:00:44.730 --> 00:00:47.420
Itβs the same symbol that we use for absolute value.
00:00:47.940 --> 00:00:52.960
For a two-by-two matrix, itβs given by the formula ππ minus ππ.
00:00:53.570 --> 00:01:00.190
Remember, weβve got to be really careful when handling negative entries in a matrix, especially when weβre trying to find the determinant.
00:01:00.870 --> 00:01:06.350
As an example, find the determinant of the matrix five, one, negative five, five.
00:01:06.690 --> 00:01:08.370
Letβs call this matrix π΅.
00:01:08.980 --> 00:01:16.230
Then we calculate the determined of π΅ to be five times five minus one times negative five.
00:01:16.690 --> 00:01:19.960
We calculate this to be 25 minus negative five.
00:01:20.110 --> 00:01:26.260
And this is where we must be really careful with the negatives as this is 25 add five, which is 30.
00:01:27.180 --> 00:01:30.300
But remember, the entries of the matrix are not always numbers.
00:01:30.860 --> 00:01:35.560
For instance, find the determinant of the matrix πΆ equals π₯, π¦, π§, π₯.
00:01:36.080 --> 00:01:40.300
Using the same rule, this matrix has determinant π₯ squared minus π¦π§.
00:01:41.110 --> 00:01:47.850
Moving on to think about three-by-three matrices, there are specific methods for finding the determinant of a three-by-three matrix.
00:01:48.160 --> 00:01:53.620
But here weβre going to develop a more general approach so that this method can be applied to larger matrices.
00:01:54.050 --> 00:01:59.410
The first thing weβre going to look at in order to be able to calculate three-by-three determinants are matrix minors.
00:02:00.660 --> 00:02:10.870
Consider a matrix π΄ with order π by π, then the matrix minor, π΄ ππ, is the initial matrix π΄ after having removed the πth row and the πth column.
00:02:11.180 --> 00:02:16.430
This means that π΄ ππ is a matrix with order π minus one by π minus one.
00:02:17.230 --> 00:02:25.460
Recall that the order of a matrix means the size of the matrix, so an π-by-π matrix is a matrix with π rows and π columns.
00:02:25.950 --> 00:02:29.580
The easiest way to understand matrix minors is through an example.
00:02:30.170 --> 00:02:32.260
Suppose we have this matrix π΄.
00:02:33.020 --> 00:02:36.440
Letβs calculate the matrix minor π΄ two three.
00:02:37.240 --> 00:02:41.920
This means removing the second row and the third column from the initial matrix.
00:02:42.540 --> 00:02:47.720
We can see that the entries that remain are the entries negative two, three, zero, negative three.
00:02:48.210 --> 00:02:51.620
So this is the two-by-two matrix minor π΄ two three.
00:02:52.540 --> 00:02:55.730
Letβs also calculate the matrix minor π΄ three one.
00:02:56.400 --> 00:02:59.230
This means removing the third row and the first column.
00:02:59.830 --> 00:03:04.370
And we can see that weβre left with the two-by-two matrix three, negative three, three, six.
00:03:05.390 --> 00:03:10.740
Weβre going to use this concept with the formal definition of the general form of the determinant of a matrix.
00:03:11.430 --> 00:03:23.260
Consider a square matrix π΄ with order π by π, then the determinant of π΄ is calculated in one of two ways, both of which involve the determinant of particular matrix minors of π΄.
00:03:24.170 --> 00:03:27.880
We can choose to calculate the determinant by using a particular row.
00:03:28.250 --> 00:03:30.950
Alternatively, we can choose one particular column.
00:03:32.080 --> 00:03:35.460
This is quite an overwhelming definition of the determinant.
00:03:35.670 --> 00:03:39.170
And it makes a lot more sense when we see it applied to an example.
00:03:39.790 --> 00:03:45.950
Find the determinant of the matrix one, two, three, three, two, two, zero, nine, eight.
00:03:46.380 --> 00:03:48.530
Letβs label this matrix as π΄.
00:03:49.040 --> 00:03:54.130
We can calculate the determinant of π΄ by first selecting either one row or one column.
00:03:54.760 --> 00:04:00.050
A good strategy is to choose the row or column which contains the most number of zeros.
00:04:00.530 --> 00:04:07.440
For this matrix, this would either be the third row or the first column, which both contain one zero entry.
00:04:08.110 --> 00:04:09.670
Letβs choose the first column.
00:04:10.230 --> 00:04:16.180
And here is the formula for finding the determinant of a matrix when we use a particular column π.
00:04:16.720 --> 00:04:20.780
For this question, as weβre using the first column, π is equal to one.
00:04:21.150 --> 00:04:25.240
And because this is a three-by-three matrix, π runs from one to three.
00:04:25.900 --> 00:04:30.010
Here is the same formula but adapted so that π runs from one to three.
00:04:30.220 --> 00:04:33.490
And as weβre using the first column, π is equal to one.
00:04:34.180 --> 00:04:39.960
Now π one one, π two one, and π three one are the entries in the column that weβre using.
00:04:40.290 --> 00:04:43.100
These are the entries one, three, and zero.
00:04:43.480 --> 00:04:46.360
And these are the determinants of the matrix minors.
00:04:46.830 --> 00:04:54.180
Recall that we find the matrix minor π΄ one one by removing the first row and the first column of the original matrix.
00:04:54.390 --> 00:04:58.310
Weβre left with the two-by-two matrix two, two, nine, eight.
00:04:58.810 --> 00:05:05.780
We then calculate the determinant of this matrix using the usual formula for finding the determinant of a two-by-two matrix.
00:05:06.120 --> 00:05:09.240
This is two times eight minus two times nine.
00:05:09.630 --> 00:05:13.560
This is 16 minus 18, which gives us negative two.
00:05:14.120 --> 00:05:19.760
We can then calculate the matrix minor π΄ two one by removing the second row and the first column.
00:05:20.160 --> 00:05:23.200
This gives us the matrix two, three, nine, eight.
00:05:23.620 --> 00:05:28.300
We can then calculate its determinant to be two times eight minus three times nine.
00:05:28.610 --> 00:05:30.510
And we find that this is negative 11.
00:05:30.950 --> 00:05:38.910
We do the same, calculating the matrix minor π΄ three one by removing the third row and the first column of the original matrix.
00:05:39.700 --> 00:05:45.750
And we find its determinant to be two times two minus three times two, which gives us negative two.
00:05:46.280 --> 00:05:55.140
The final step is to calculate the values negative one to the power of one add one, negative one to the power of two add one, and negative one to the power of three add one.
00:05:55.580 --> 00:05:58.440
We find these to be one, negative one, and one.
00:05:58.940 --> 00:06:05.410
Remember, raising negative one to an even power gives us one and raising negative one to an odd power gives us negative one.
00:06:05.900 --> 00:06:09.860
Letβs now substitute into this formula the values that we found.
00:06:10.240 --> 00:06:16.670
Substituting in these values becomes clear why itβs useful to choose a row or column that has the most number of zeros.
00:06:16.940 --> 00:06:25.570
Because this last term is being multiplied by zero, the whole term will be zero, which just means itβs one less thing that we actually need to calculate.
00:06:26.180 --> 00:06:35.460
Tidying things up a bit, we find that one times one times negative two gives us negative two and negative one times three times negative 11 gives us 33.
00:06:35.740 --> 00:06:39.850
Adding these together, we get the final answer of 31.
00:06:41.210 --> 00:06:49.530
So the main things to think about are which row or column youβre going to be using and carefully calculating each little bit of the formula.
00:06:50.090 --> 00:06:53.800
And weβve got to make sure that weβre always really careful around negatives.
00:06:54.560 --> 00:06:59.840
Letβs now see another example, although, this time, weβll expand along a row rather than a column.
00:07:00.830 --> 00:07:08.180
Calculate the determinant of π΄ when π΄ equals three, zero, negative one, zero, one, zero, two, two, four.
00:07:09.050 --> 00:07:15.580
To make things easier for ourselves, weβll identify the row or column that contains the most number of entries which are zero.
00:07:16.150 --> 00:07:19.830
For this matrix, thatβs going to be the second row of π΄.
00:07:20.470 --> 00:07:26.000
Letβs remind ourselves of the general formula for finding the determinant of an π-by-π matrix.
00:07:26.760 --> 00:07:27.940
Here is the formula.
00:07:28.260 --> 00:07:32.810
Remember that π represents the row number and π represents the column number.
00:07:33.670 --> 00:07:37.270
So as we have three columns, π runs from one to three.
00:07:37.680 --> 00:07:41.100
And weβre expanding along the second row, so π equals two.
00:07:41.640 --> 00:07:44.230
So this is the formula that weβre going to be using.
00:07:44.940 --> 00:07:51.500
The entries π two one, π two two, and π two three are zero, one, and zero, respectively.
00:07:51.970 --> 00:07:59.520
Because two of these entries are zero, we find that both of these terms are going to be zero because theyβre both being multiplied by zero.
00:08:00.000 --> 00:08:03.170
So thereβs actually only one term that we need to calculate here.
00:08:03.910 --> 00:08:07.160
Letβs begin by calculating the matrix minor π΄ two two.
00:08:07.790 --> 00:08:13.050
This is going to be the two-by-two matrix we get by removing the second row and the second column.
00:08:13.720 --> 00:08:19.520
So we see that the matrix minor π΄ two two is three, negative one, two, four.
00:08:20.240 --> 00:08:23.680
But what we actually need for our formula is its determinant.
00:08:24.110 --> 00:08:29.140
So we calculate this in the usual way for finding determinants of a two-by-two matrix.
00:08:29.760 --> 00:08:33.960
This gives us three times four minus negative one times two.
00:08:34.530 --> 00:08:38.900
This gives us 12 minus negative two, which gives us 14.
00:08:39.850 --> 00:08:44.380
The final thing we can do is calculate the value of negative one to the power of two add two.
00:08:44.820 --> 00:08:51.380
As this is negative one to the fourth power, which is an even power, this is going to give us one.
00:08:52.070 --> 00:08:56.510
So we calculate the determinant of π΄ to be one times one times 14.
00:08:57.210 --> 00:08:59.640
And this, of course, just gives us 14.
00:09:00.480 --> 00:09:09.610
So carefully selecting the row or column that you choose when youβre finding the determinant of a matrix can really reduce the amount of calculations necessary.
00:09:11.060 --> 00:09:14.800
Remember that not all matrices contain numeric entries.
00:09:15.920 --> 00:09:17.930
Letβs have a look at this in an example.
00:09:19.200 --> 00:09:24.730
Consider the determinant π₯, π§, π¦, π¦, π₯, π§, π§, π¦, π₯.
00:09:25.150 --> 00:09:34.970
Given that π₯ cubed add π¦ cubed add π§ cubed equals negative 73 and π₯π¦π§ equals negative eight, determine the determinantβs numerical value.
00:09:35.960 --> 00:09:39.930
Remember that we can choose to calculate this determinant by picking a row or column.
00:09:40.320 --> 00:09:43.270
We often choose the row or column with the most amount of zeros.
00:09:43.480 --> 00:09:47.870
But as we have three unknowns, π₯, π¦, and π§, letβs just choose the top row.
00:09:48.590 --> 00:09:55.820
And here is the general formula to find the determinant of an π-by-π matrix, using a particular row, π.
00:09:56.610 --> 00:10:00.840
As weβre using a three-by-three matrix, π runs from one to three.
00:10:02.000 --> 00:10:05.490
And as weβre using the top row, π is equal to one.
00:10:06.270 --> 00:10:10.950
So here is the particular version of this general formula applied for our question.
00:10:11.430 --> 00:10:13.590
Letβs clear some space before we continue.
00:10:14.310 --> 00:10:16.440
Iβve kept the formula that weβre using on the screen.
00:10:17.370 --> 00:10:19.750
Letβs call the matrix that weβre using π΄.
00:10:20.450 --> 00:10:24.490
The straight lines either side of the matrix denote the determinant.
00:10:25.420 --> 00:10:29.510
Letβs begin by finding the different components in the formula that weβre using.
00:10:29.960 --> 00:10:36.030
The lowercase π one one, π one two, and π one three are the entries in the row that weβre using.
00:10:36.430 --> 00:10:41.100
So π one one is π₯, π one two is π§, and π one three is π¦.
00:10:41.880 --> 00:10:47.560
The uppercase, π΄ one one, π΄ one two, and π΄ one three denote the matrix minors.
00:10:48.390 --> 00:10:55.050
Remember that we get these from removing a particular row and a particular column from the original matrix.
00:10:55.750 --> 00:11:01.520
For example, we get π΄ one one by removing the first row and the first column, like so.
00:11:01.550 --> 00:11:05.670
And weβre left with the two-by-two matrix π₯, π§, π¦, π₯.
00:11:06.270 --> 00:11:08.470
We then do the same for π΄ one two.
00:11:08.890 --> 00:11:13.000
We get the matrix minor by removing the first row and the second column.
00:11:13.420 --> 00:11:16.390
This leaves us with the two-by-two matrix π¦, π§, π§, π₯.
00:11:16.390 --> 00:11:23.060
And finally, we get the matrix minor π΄ one three by removing the first row and the third column.
00:11:23.420 --> 00:11:26.970
This leaves us with the two-by-two matrix π¦, π₯, π§, π¦.
00:11:27.990 --> 00:11:32.650
But what we actually need for our formula is the determinant of each of these matrices.
00:11:33.230 --> 00:11:36.920
Letβs recall a method to find the determinant of a two-by-two matrix.
00:11:37.500 --> 00:11:42.820
So the determinant of the matrix minor π΄ one one is π₯ times π₯ minus π§ times π¦.
00:11:43.870 --> 00:11:47.490
We can equivalently write this as π₯ squared minus π¦π§.
00:11:48.430 --> 00:11:57.740
In the same way, we find the determinant of the matrix minor π΄ one two to be π¦ times π₯ minus π§ times π§, or equivalently π₯π¦ minus π§ squared.
00:11:58.390 --> 00:12:06.880
And we find the determinant of the matrix minor π΄ one three to be π¦ times π¦ minus π₯ times π§, which is equivalently π¦ squared minus π₯π§.
00:12:07.620 --> 00:12:17.450
The final thing we need to calculate for our formula are the values of negative one to the power of one add one, negative one to the power of one add two, and negative one to the power of one add three.
00:12:17.900 --> 00:12:24.740
Remember that raising negative one to an even power gives us one and raising negative one to an odd power gives us negative one.
00:12:25.110 --> 00:12:29.220
So negative one to the power of one add one is negative one squared, which is one.
00:12:29.510 --> 00:12:33.930
Negative one to the power of one add two is negative one cubed, which is negative one.
00:12:34.210 --> 00:12:39.770
And finally, negative one to the power of one add three is negative one to the fourth power, which gives us one.
00:12:40.260 --> 00:12:44.700
So letβs not write out the determinants of this matrix with the components that we found.
00:12:45.210 --> 00:12:48.050
From here, letβs just try and simplify a little bit.
00:12:48.650 --> 00:12:52.950
Letβs begin by multiplying together the brackets with algebraic terms.
00:12:53.310 --> 00:12:56.900
For the first one, we have π₯ multiplied by π₯ squared minus π¦π§.
00:12:56.900 --> 00:13:01.890
So this term becomes one times π₯ cubed minus π₯π¦π§.
00:13:02.640 --> 00:13:04.440
We then do the same with the next term.
00:13:04.640 --> 00:13:09.110
This gives us negative one times π₯π¦π§ minus π§ cubed.
00:13:09.710 --> 00:13:15.500
And repeating the same with the last term, we end up with one times π¦ cubed minus π₯π¦π§.
00:13:15.500 --> 00:13:20.330
So now, weβve multiplied together the brackets containing algebraic terms.
00:13:20.600 --> 00:13:23.790
Letβs multiply each term by the value at the front.
00:13:24.030 --> 00:13:30.830
The first term is one times π₯ cubed minus π₯π¦π§, which is of course going to give us π₯ cubed minus π₯π¦π§.
00:13:30.830 --> 00:13:36.540
The next term is negative one times π₯π¦π§ minus π§ cubed.
00:13:37.050 --> 00:13:40.060
So we can multiply the second bracket through by negative one.
00:13:40.320 --> 00:13:46.740
But weβve got to be careful because the π§ cubed already has a minus at the front of it, so thatβs going to become a positive.
00:13:46.980 --> 00:13:50.260
So weβre gonna end up with minus π₯π¦π§ add π§ cubed.
00:13:50.490 --> 00:13:53.970
Finally, the last term is just one times π¦ cubed minus π₯π¦π§.
00:13:53.970 --> 00:13:57.270
So this gives us π¦ cubed minus π₯π¦π§.
00:13:58.220 --> 00:14:00.130
We can now simplify this a little bit.
00:14:00.500 --> 00:14:03.350
We can gather together the π₯π¦π§ terms.
00:14:03.800 --> 00:14:06.860
Bringing these together gives us minus three π₯π¦π§.
00:14:07.770 --> 00:14:12.780
From here, we notice that we have π₯ cubed add π§ cubed add π¦ cubed.
00:14:13.050 --> 00:14:18.100
Weβre told in the question that π₯ cubed add π¦ cubed add π§ cubed gives us negative 73.
00:14:18.670 --> 00:14:26.210
Additionally, we have negative three π₯π¦π§, and weβre told in the question that π₯π¦π§ is equal to negative eight.
00:14:26.630 --> 00:14:34.910
So substituting in those values gives us negative 73 minus three times negative eight, which is negative 73 add 24.
00:14:35.150 --> 00:14:37.400
And this gives us negative 49.
00:14:38.560 --> 00:14:46.020
So sometimes, we encounter matrices which donβt have numerical values, but we can still find the determinant in exactly the same way.
00:14:46.710 --> 00:14:49.380
Letβs now summarize the main points from this lesson.
00:14:50.240 --> 00:14:58.240
For a matrix π΄, a matrix minor π΄ ππ is the initial matrix π΄ after having removed the πth row and the πth column.
00:14:58.890 --> 00:15:03.550
And here is the general formula for the determinant of an π-by-π matrix π΄.
00:15:04.230 --> 00:15:07.350
We can either choose a particular row or a particular column.
00:15:07.980 --> 00:15:12.900
But it often helps to choose a row or column that contains the most number of zeros.