00:01
We are going to solve the linear equation system given below for four steps with zero error, epsilon and equal zero.
00:11
Isn't the gosatal iteration method? and the system is 4x1 plus x2 equal 1, x1 plus 4x2 minus x3 equal 2 and x2 plus 4x3 equal 3.
00:27
So i'm going to write that linear system here in matrix vector 4x2.
00:33
The coefficient matrix is 410 first row then 1 -9 negative 1 and 0 1.
00:48
Times the unknown vector is x1 x2 x3 equal the right side is 1 2 2 3.
01:05
So if we call this a, coefficient matrix, b, the right hand side and x, the unknown vector, we have written the linear system in a matrix vector form like this.
01:22
And now we can partition this matrix this way.
01:30
If we define here, we say a equal d minus e minus f.
01:44
Where d is a diagonal matrix with main diagonal equal to the main diagonal of a that is 4 -00, 0 -40, and 0 -0 -0 -0 -4.
01:59
Then e is the lower triangle of a without the main diagonal, changing the sign of all elements.
02:09
So we get 0 -0, 0 ,000, and at the 1 -0 .0.
02:20
And 0 negative 1, 0.
02:26
And f is the upper triangular of a without the main diagonal and changing the sign of all the elements.
02:33
That is, f is 0, negative 1, 0, 0, 0, 0, 0.
02:41
0 ,0.
02:46
It is clear that we have partition a this way.
02:51
D -a minus e minus f for the e -e -e -e -m -a.
02:54
And f equal to these matrices.
02:59
So having done this, we can say that a is equal to, we can group together the first two matrices, d minus e, minus f, which becomes, let's say, m minus n, where m is the matrix d minus e, and n is f.
03:25
And it is clear that m is the lower, triangle of a that is m is 4 -0 -0 -1 -4 -0 and 0 -1 -4 and n is f that is n is 0 negative 1 -0 -0 -0 -0 here and 0 -0 so the the gosado method use this partition of a given here.
04:22
So gosadl corresponds to this partition of a.
04:41
And we see here that a x equal b implies that we have written a as m minus n times x equal b.
04:54
And this is equivalent, in fact, is equivalent to mx.
05:01
Equal n x plus b and from this same equivalent system we write the iteration method so the ghost saddle iteration reads m x k plus one we are going to use supper indices between parentheses to denote the iterate vectors because we are going to use the sub indices for the components of the vectors.
05:40
So mx iterate k plus 1 is equal to n iterate x x k plus b.
05:50
This is the gosato duration.
05:56
And that's for k grade integral 0.
05:58
That is we need an initial vector x0 to start the iterations.
06:09
So we have that and in this case, this method is going to converge for any initial vector because the matrix a is diagonal dominant.
06:30
So we can say that.
06:32
A is diagonal dominant.
06:42
Remember, that means that the absolute value of the elements on the diagonal are than the sum of the off diagonal elements in absolute volume and that true for the whole all the rows of the matrix so diagonal dominant implies cos sato converge for any initial vector and that's important because we can start with zero vector which is maybe the easier one so we start with that vector x not equal zero zero zero and with that we that we calculate x1 using this iteration here so mx1 equal n x0 plus b but x not is 0 it means this is 0 and this is b so the linear system we get a solved is 4 0 140 1 410 1 410 0 1 414 times the another vector will be the iterated of index 1, component 1, iterate 1, component 2, and iterate 1, iterate 1, component 3.
08:12
And that's equal to b is 1 to 3.
08:17
And this can be solved, and we are going to say it only here, and we know all the other iterates are calculated with the same procedure that is for substitution.
08:32
That is because the coefficient, matrix of this system here is a lower triangular matrix so we can solve it at four substitution that is we start with the first equation and proceed down to the last equation and at each step we are calculating each component of the unknown vector so this is the method we're going to use each time we got to solve the equation given by this iterate here it's always going to be forward substitution because the matrix m is constant.
09:13
Remember, this is the characteristic of the stationary iterative method to solve linear system.
09:20
Okay, so we have this and we can say that x1 is directly one -fourth.
09:30
Then we can solve the second equation to say that x21 is 2 minus x11.
09:41
Over 4 that is 2 minus 1 fourth over 4 and that is 7 over 16 and the third equation and give us x 3 x1 3 is 3 minus x2 or x12 over 4 and that is 3 minus the previous coordinate this one here 7 over 16 over 4 and that is equal to 41 over 64.
10:28
Okay, so we have calculated vector x1, this next iterate of the gos -satom method, and that is equal to vector 1 4th, 7 over 16, and 41 over 64 transpose.
10:48
That is because we are taking always the vectors to be vertical, that is, matrices we want column.
10:58
Okay, so this is x1, and now we calculate the next iterate, x2.
11:03
We use the same formula, remember here.
11:08
Now, in this case, for k equal 1, we get mx2.
11:15
Let's, sorry, let's say we are calculating now an exiterate is mx2 equal nx1 plus v.
11:32
M same matrix 4 zero one four zero and zero and four times the unknown vector is iterate two coordinate one a three coordinate two coordinate two it rate two coordinate three equal n is the metric zero negative one zero zero zero 01 and 000 times the previous seat rate here 1 4th 716th and 41 or 64 plus b 1 2 and 3 we do all the this operation here it's very easy is the new system is 4 001 140 1 1401 1 1 4 times x1 .2 or x21, x2 and x23 equal, and here is negative 7 over 16, 41 over 64, and the third one is 0, plus 1, 2, and 3.
13:04
And this becomes finally vector 9 over 16, 169, sorry, 169 over 64 and 3.
13:23
Okay, now we solve this lower triangular system by back, forward substitution, sorry.
13:30
And then we get first x1 .2 is one -fourth times 9 over 16.
13:42
And that is 9 over 64.
13:47
Then the second component of the second iterate, or the third iterate, if you want, is the fraction 169 over 64 minus x2 .1 over 4, and that is fraction 169 over 64...