00:01
In this question we have been given a system of equation.
00:04
We need to solve the system using the goss siddell method.
00:08
Okay.
00:09
And we need to solve it until the percent relative error falls below 5 percent.
00:16
Okay.
00:17
So, and we need to must check whether this equation satisfy this relation or not.
00:23
That is, these sum, modulus of aii, i, is greater than some.
00:30
Of mod of a .i .j.
00:34
So what does this mean? first, let me see that.
00:37
Okay, relaxation is given as lambda equal to 1 .2.
00:41
Okay.
00:43
So with the help of this system, we can get a matrix.
00:47
So since we know that ax equal to b.
00:51
So we will try to first write down what is the matrix a, that is a coefficient matrix.
00:57
So 3, 6, minus 2, then 10, 2 minus 1 and what else we will get 1 1 1 5 1 and 5 so this is the matrix that we are getting in correct if we interchange r1 and r2 so we will interchange r1 and r2 so after interchange the two rows what i will get i will get 10 to minus 1 and here i will get 3 6 minus 1 and here i will get 3 6 minus 2 and this is 1 1 at 5 okay now what we can do we can clearly check so you see 10 10 is greater than 3 plus 2 sorry 10 is greater than 2 plus 1 so which is obvious okay so that is equal to 3 then 6 you can see that it is greater than 3 plus 2 so which is 5 only again 5 is greater than 1 plus 1 which is again do.
02:04
Okay, so this condition is satisfied here.
02:08
So we say that.
02:10
So this implies that the cost -siddle method is convergent.
02:16
Okay, so this implies that the cost -siddle method, saddle -method is convergent, correct? so now we need to apply the method, correct? so we have from the given system from the first equation, i will write down the value of x1.
02:38
So it will be 1 divided by 10, 27 minus 2 x2 plus x3.
02:45
Similarly, i would try to find out x2, so it will be 61 .5 minus 3 x1 plus 2 x3 divided by 6.
02:56
Then x3 is equal to.
02:59
From the given equation, i'm just evaluating this x1, x2, x3.
03:03
Okay, nothing else i'm doing here minus x2 times one divided by five now we have been given lambda as a relaxation okay so we know that the iteration which is denoted by x.
03:23
New is equals to it equal to lambda times of x.
03:31
J new correct plus one minus lambda times of x j hold so this is the iteration okay now we need to choose the initial iteration so our initial iteration will be initial iteration okay that we will choose as x1 x2 zero and x3 0 as 0 comma 0 comma 0 correct so what will be the value of x1 then in that case if i try to point out the value of x1 in the first iteration it will be just from this equation from equation 1 okay then from equation 1 x2 x3 put equal to 0 then x 1 will be what 27 divided by 10 which is nothing but 2 .7 now we know the value of x1.
04:36
We can calculate the value of x2.
04:38
So x2 in the first iteration, it will be 1 divided by 6 .61 .5 minus 3x1 plus 2 times of x3 at the 0 iteration.
04:52
Okay.
04:53
Now we know the value of x1 as well.
04:55
So 1 divided by 6, 61 .5 minus 3 times of 2 .7.
05:01
So which is clearly 8 .9.
05:05
Correct.
05:06
Similarly, for x3 also i can find it out.
05:10
So x in the first separation, it will be now we know x1 and x2.
05:14
So it will be minus 21 .5 minus x1 in the first iteration, minus x2 in the first iteration.
05:23
Okay, so it comes out to be minus 6 .62...