00:01
Hi, now we're given the equations here, 2x plus 2y, 2x plus 2y minus z, that equals 2.
00:11
That's one equation.
00:12
Second is x minus 3y plus c is equal to minus 28.
00:22
So we have, next one is minus x plus y plus 0, 0, 0, 0, and that equals 14, right? so we're going to solve this equations here using the goals -jordan elimination method.
00:41
So for that, what we'll do, we'll make here the augmented matrix 3 cross 4, right? so, now we first put the coefficients here, there'll be 2 to minus 1, 1 minus 3 1, and next we put minus 1, 1, 0.
01:01
Let's report along with that, we put the constants here, that is given as 2, minus 28, and it is given as 14 here, right? so this we got here.
01:16
Now, the purpose is here, we need to make this on the unit matrix by applying raw operations and then whatever we get here, that will be solutions.
01:25
I'm again debating, we make this as a unit matrix, right? by row operations and whatever we get here, xyz, that will give the solutions here.
01:34
So let's go for that.
01:36
So let's try to make this as a unit matrix.
01:39
First, what we can do here, you make this as a one basically.
01:42
So apply the row operation r1 tends to r1 by 2.
01:47
Right? this we get.
01:49
So here this will be connected to 1.
01:52
1 minus 1 over 2.
01:57
Then we get this will be 1.
01:58
R2r3 as it is.
02:00
It is 1 minus 3, 1 minus 28 minus 1, 1, 0, it is 14.
02:10
Right, this we get here.
02:12
The next part is we have to make here, the next part is we can make this as 0 and this as 0.
02:22
So for that, we apply the row operation that is r2 tends to r2 minus r1 and r3 tends to r3 plus r1.
02:33
So we get the matrix, so r1 is as it is given as 1, 1 minus 1 by 2, there will be 1 here.
02:44
And then we have r2.
02:45
That is given as 0 and then minus 4 will be 1 plus 1 by 2 3 over 2 and we get minus 29 this will be 0 it will be 2 here minus 1 over 2 that's given as 15 right we get this as here 15 okay moving further now the next part is what we can do here we'll make this as 1 to make it as a unit matrix so apply the operation r2 tends to r2 over minus 4.
03:22
So we get the matrix here.
03:24
R1 as it is 1 1 minus 1 over 2, it will be 1.
03:30
Now this will be 1 here, 3 over 2 over minus 4 will be minus 3 over 8.
03:37
And that will give 9 over 4, right? now r3 as it is 0, 2, minus 1 over 2, it's given as 15, all right? next, now we can make this as zero here and this has zero.
03:55
So for that, we apply the operations that is given as r1 tends to r1 minus r2.
04:01
It will become zero.
04:03
And then r3 tends to r3 minus 2 r2...
