# Gauss Jacobi’s Method with Example – System of Linear Equations – Engineering Mathematics 1

Hello Friends, So in this video we are gonna to learn a new concept called as Jacob is iteration method now what is usual iteration method and what exactly Jacob is iteration method so let me tell you that whenever we have simultaneous linear equations and if you want to find out the unknowns then there are many methods available so you must have done the method of Crammer’s rule or the method of Gauss elimination the method of Gauss gorden so these are the some examples or some methods by which you can solve the simultaneous linear equations so Jacob is a iterative method is one of them by using this method you can solve simultaneous linear equations now now why do we call it as I try to method so the reason is in Jacobi’s or in any iterations method when we go through multiple iterations we reach near to root so in high creative methods we generally go through multiple iterations to reach the nearest root or the nearest value of the unknown now what is Jacob is I try to method so to understand it let us consider some equations so here I will say let a 1 X plus B 1 1 plus C 1 Z is equal to D 1 a 2 X plus B 2 y plus C 2 Z is equal to D 2 and a 3 X plus B 3 y plus C 3 Z is equal to D 3 are the three simultaneous linear equations now to get the value of unknowns XY and z we are going to use the Jacobi’s iterative method so generally if you observe such equations and if you want to solve it by j a trading method then you will find that the value of a 1 B 2 and C 3 are generally greater or large as compared to other coefficients of the equation so a 1 B 2 and C 3 will be always large in value as compared to B 1 C 1 a 2 C 2 a 3 B 3 or D 1 D 2 D 3 so so first of all we will identify these large coefficients so now since a 1 is the largest coefficient as well as B 2 and C 3 we will convert this first equation so to solve these equations by Jacob is iterative method we will rewrite this equation in such a form that these coefficients a 1 B 2 and C 3 will come in the denominator so here I will say that first equation I will represent as a 1 X or X is equal to I will send this on the other side so that will become D 1 minus B 1 Y minus C 1 Z and this a 1 will come in the denominator next to take B 2 in the denominator I will represent this as D 2 minus a 2 X minus C 2 Z into 1 upon B 2 so this is value of y and to take C 3 in the denominator will represent this as d 3 minus a 3 X minus B 3 y 1 upon C 3 that is the value of Z so here we got the value of x y&z now in I try to method as I say that we have to perform multiple iterations so whenever we are starting with first iteration then we have to assume certain routes so here we assume that the roots of the equation or the values of this unknowns are zero zero zero so in third step we will assume that X 0 is 0 y 0 is 0 and Z 0 is also 0 so here I will say step number 3 so by assuming the initial roots are 0 we will start dissolution and we will find the value of XY and Z and we will call it as x1 y1 and z1 so here I will say put this and find x1 y1 z1 then once we get develop x1 y1 z1 we will use it again in the same equations and we’ll find out the value of x2 y2 and z2 so that will be called as second iteration then we’ll take those values of x2 y2 z2 in the same equation to get the value of x3 y3 and z3 and we’ll continue this process till we get desired level of accuracy so once we get the desired level of accuracy we stop the operation and whatever is the value of x y&z in that I tration will be our final value of this variables XY and Z so step number four year L say continue step number three till we get desired accuracy now guys what is this desire accuracy so we generally say that whenever values of consecutive iterations are same then we got the desired accuracy so let’s say in seventh iteration you’re getting X 7 y 7 and z 7 and in the eighth iteration the value of x 8 y and z 8 are similar so if X 7 is same as exit y 7 same as Y it and that 7 is same as Z then we say that we have reached the desired accuracy and then we stop the operation and whatever is the value of x 8 y 8 and 0 that is our value of XY and Z so now we will apply this Jacob is I try to method in one example so I have an example solve following equations by Jacob is hydration method now you can see that there are three equations three unknowns and out of that this a 1 B 2 and C 3 are large as compared to other coefficients now what we’ll do is as I said in step number one we’ll convert this first equation in terms of X so this will be X equal to 1 upon 15 in bracket 18 minus 2y minus Z next y is equal to 1 by 2019 minus 2x plus 3 Z next Z is equal to 1 upon 25 22 minus 3x + 6 1 so now we got the value of x y&z now we’ll start with the iteration so for first iteration as I said that we will start by assuming the initial root as 0 0 and 0 so here I’ll say first iteration put X 0 as 0 y 0 as 0 and Z 0 as 0 so once you put these 3 values here then we get this two terms as 0 and here we get X 1 as 18 upon 15 similarly y 1 as 19 upon 20 and z1 s 22 open twenty five so here I will say x1 is 18 upon 15 y1 y1 is equal to 19 upon 20 and z1 is equal to 22 upon 25 now if will calculate this values in calculator then you will get so now after the first iteration we go develop x-one y-one and z-one now we will use these routes for second iteration so here I will say the second iteration put x1 is equal to 1.2 y1 is equal to 0.95 and z1 is equal to 0.88 so if we put these equations here we’ll get X 2 as 1 by 15 18 minus 2y 1 minus Z 1 so here I will say therefore x2 is equal to 1 upon 15 18 minus 2 times y 1 so Y 1 is zero point nine five minus zero point eight 8 y 2 will be 1 upon 20 19 minus 2 X 1 plus 3 Z 1 1 upon 20 19 minus 2 X 1 is 1 point 2 plus 3 Z 1 that is 0.88 and Z 2 will be 1 by 25 22 minus 3 X 1 plus 6y 1 so if you will solve this in calculator then you will get x2 as + y-two and z-two s now here we got the values of x2 y2 and z2 we’ll put these values in the third iteration to get the value of x3 y3 and z3 so here I’ll say you you and if you will solve these three equations in calculator then you will get an F you’ll solve these three equations in calculator then you will get now we’ll put these values of x3 y3 z3 in fourth iteration to get the value of x four by four z4 and we’ll continue this process till we get the desired level of accuracy so the fourth iteration will give us a false forward ahsoka a fast forward car than a jovially corona be now we’ll solve these three equations in calculator then you will get the values of X 4 y 4 and 0 for s so now we will put this values of x4 y4 + z4 in next iteration that is tooth titration so we will get so it will solve these two equations kelsey then we will get now we’ll use these values of x5 by fans I find the next iteration that is sixth iteration and we’ll get x 6y 6 + z 6 you so if we’ll put these values in the calculator then we will get now if you will see the fifth and sixth iteration then you can see that the value of x-5 x6 is going towards 1 so in the next iteration we will get value as 1 or here also it is almost equal to 1 next the value of y 5 is also going towards one because 0.999 for it we are getting 9 9 9 6 so it will be 1 and similarly value of z is going towards 1 that is zero point nine nine eight zero zero point nine and nine eight so in next iteration we will get the values of 1 1 1 and therefore I will say here that the value of x is equal to value of y is equal to value of Z that is equal to 1 so here we got the values of x y and z by using Jacobi’s iteration method thank you