00:01
All right, so first thing we want to try to do with solving a system of equations using the gauss -jordan elimination method is we're trying to take our system and write it in terms of an augmented matrix.
00:17
So we're going to create that augmented matrix by capturing that equation.
00:25
So the first linear equation using coefficients, the coefficient of x is negative 1.
00:30
The coefficient on y is 1 and then the solution to that equation is negative 22 right and then we're going to move on to the second equation the coefficient on x is 3 the coefficient of y is 4 and the third equation coefficient on x is 4 coefficient on y then the solution is 32 all right so now when we're using the gouse jordan elimination method what we're what we want to try to do is we want to this row to become ones and then we want to zero everything else above and below it.
01:09
All right? so one of the first things we can do to get that first row to become, that first negative one to become positive one is we simply take the first row and multiply it by a negative one.
01:21
So negative one times a one.
01:25
And we get the first row negative one times negative one to one.
01:31
Negative 1 times 1 is in the heat of 1 then negative 1 times negative 22 and positive 22 okay so what i'll do is i'll copy the next two rows down so this will be a 3 and then that's 4 4 8 i'm not changing anything i'm just copying it down so we can just do one step at a time here all right so now what i want to and my next is i want this to be a zero and i want this to be a zero.
02:11
And i'm going to use this one in this first row.
02:15
And i'm going to use row operations with matrices to try to get these to be zeros.
02:20
All right.
02:20
So what we're going to do is we're going to do two steps here.
02:24
First thing i'm going to do is i'm going to take row one and multiply it by a negative three.
02:28
So negative three times row one.
02:32
And we're going to add that to row two.
02:36
Right and then i'm going to take row one and multiply it by negative four so negative four times row one i'm going to add that to row okay so let's see what we get when we do that all right so row one is fixed one negative one and then we're going to get that 22 so here we go negative three times one is a negative three and three is zero that's exactly what i want to get there negative three times a negative one is a positive three and four is seven and then negative three times 22 is a negative 66 and 4 is a negative 60 right negative 4 times 1 now using this here this one here so we're going to move down to here now right so negative 4 times or 1 that's 1 and 0 and 4 becomes 0 negative 4 times a negative 1 becomes a positive 4 and a negative 8 that is going to become a negative 4 and then a negative 4 times 22 is a negative 88 and a negative 88 and 32 is a negative 50.
03:59
Okay.
04:00
So next.
04:05
Let's see if i can squeeze this up in here.
04:07
So what we're going to do now is we are going to now try to get this to become a 1 and that to become a 1.
04:26
We can do that all in the same step.
04:28
We can take this second row and divide it by 7.
04:31
So i'm going to take one seventh row two and i want to take this and get this to be a one.
04:40
So i'm going to take this and divide everything here by four or take a negative one fourth of row.
04:50
Okay.
04:51
So row one stays the same.
04:53
1, negative 1 and 22.
04:58
Row 2, negative 1 7 of 0 is 0.
05:03
Negative 1 7th of 7.
05:04
Or one -seventh, i should say, of seven is one, and then one -seventh of a negative 62 is a negative 62 over seven.
05:14
All right, and the negative one -fourth of zero is zero.
05:17
A negative -fourth of a negative four is a positive one.
05:21
And a negative -fourth of a negative 56, that's going to turn out.
05:25
All right, so now i'm going to take this that i have here, or here, i'm going to bring it down and rewrite it.
05:35
So we can pick up the problem...