00:01
In this video, we'll be finding the rref, or the row reduced epsilon form of this system of linear equations and providing a solution to the system.
00:10
So the first step in handling a question like this is we convert this system into a matrix, an augmented matrix.
00:22
So all we do is we look at the coefficients.
00:28
So here we have 1, here's 2, 1, 1, and of course 1 on the right as the constant value.
00:35
1, 2, 4, negative 3, 2, negative 3, 3, 6, negative 3, 3 and negative 3.
00:45
And we write them in our matrix.
00:47
So we just go along.
00:49
So 1, 2, 1, 1, 1.
00:54
Oops, went over here.
00:57
So that's the first row.
00:59
And as you notice, i'm just using the values underlined in red.
01:02
So here's the second row.
01:05
2, 4, negative 3, 2, and negative 3.
01:13
And the final row.
01:14
3, 6, negative 3, 3, negative 3.
01:21
Okay, now we can go about solving the system.
01:26
So a good first step, you can notice that we can have row 2 equals row 2 plus negative 2 times row 1.
01:39
So we can write 2, 4, negative 3, 2, negative 3, which is row 2, plus and then just negative 2 times all the values in row 1.
02:00
So negative 2 times 1 is negative 2.
02:03
Negative 2 times 2 is negative 4.
02:06
Negative 2 times 1 is negative 2, negative 2, negative 2.
02:12
So now we just add these expressions or add these rows.
02:17
So 2 plus negative 2, 0.
02:19
4 plus negative 4, 0.
02:23
Negative 3 plus negative 2 is negative 5.
02:26
2 plus negative 2 is again 0.
02:29
And negative 3 plus negative 2 is negative 5.
02:32
And if you notice, we have negative 5 for negative 5.
02:36
We can just do a shortcut here and multiply this whole row by negative 1 over 5 to cancel them out.
02:42
And we can see that we get a nice 0, 0, 1, 0, 1.
02:47
Which we can then substitute into our augmented matrix as our new row 2.
02:55
Before doing that though, we can see that we can also change row 3 by doing a similar operation.
03:03
So we do row 3 equals row 3 plus a negative 3.
03:12
Or i guess we can do minus 3 times row 1.
03:17
So first you write row 3, which is 3, 6, negative 3, 3, negative 3.
03:28
Plus, and then again like we did before, negative 3 times every value in row 1.
03:37
So negative 3 times 1 is negative 3.
03:41
Negative 3 times 2 is negative 6.
03:44
Negative 3 times 1 again is negative 3, negative 3, and negative 3.
03:52
Okay, so on the bottom, 3 plus negative 3 is 0.
03:55
Let's write that in green.
03:56
Okay, 3 plus negative 3 is 0.
04:00
6 plus negative 6 is 0.
04:03
Negative 3 minus 3 is negative 6.
04:06
3 plus negative 3 is 0.
04:08
Negative 3 minus 3 is negative 6.
04:11
A similar operation, multiply this whole row by negative 1 over 6 to cancel out the negative 6.
04:17
And we get 0, 0, 1, 0, 1.
04:22
And as you notice, this is the same as row 2.
04:26
So now let's get these values and write them in our matrix.
04:33
So row 1 is unchanged as of yet.
04:36
So we can still write 1, 2, 1, 1, 1.
04:40
And now we write row 2, which is in purple.
04:45
0, 0, 1, 0, 1.
04:48
And now the value is in green.
04:51
0, 0, 1, 0, 1.
04:54
Now we can simply see that row 2 and row 3 are equal.
05:00
So what we can do is row 3 equals row 3 minus row 2...