 नमस्ते, मैशत दक्तर बसरज मबरादर, हूल्चट ईनिस्ट्रूट तकनलोज सोलापुर, अवसे सी लिन्यदा आजभीक एकवीशन बाई, जोग्बिश इट्रीशन मेथाः. that is the first iteration of y 1 is equal to 1 upon a 22 into b 2 minus a 21 into x 0 minus a 23 into z 0 z 1 is equal to 1 upon a 33 into b 3 minus a 31 into x 0 minus a 32 into y 0 call it is equation number 3 substituting these value of x 1 y 1 z 1 and attend up equation 2 the second iteration is given by x 2 y 2 and z 2 that is the number of iteration or approximation that is x 2 is equal to 1 upon a b 1 minus a 12 into y 1 minus a 13 into z 1 y 2 is equal to 1 upon a 22 into b 2 minus a 21 into x 1 minus a 23 into z 1 z 2 is equal to 1 upon a 33 into b 3 minus a 31 into x 1 minus a 32 into y 1 proceeding in the same way we get the successive iteration in general the case iteration is x suffix k is equal to 1 upon a 11 into b 1 minus a 12 into y suffix k minus 1 minus a 13 into z suffix k minus 1 y k is equal to 1 upon a 22 into b 2 minus a 21 into x suffix k minus 1 minus a 23 into z suffix k minus 1 and z suffix k is equal to the 1 upon a 33 into b 3 minus a 31 into z suffix k minus 1 minus a 32 into y suffix k minus 1 the iteration process is stopped when the desired order of approximation is obtained or or two successive iteration are nearly the same the final value of x y z so obtained is called an approximate solution of x system 1 pass the video answer the following question are the following system of equations are diagonally dominant if not write a diagonally dominant form I hope all of you obtain the result here by observing the given system of equation the coefficient matrix of the given system of equation is not diagonally dominant dominant you rearrange the given system of equation such a way that the coefficient of x is larger than the coefficient of y and z in that equation and coefficient of y is larger than the coefficient of x and z in the second equation and the coefficient of z is larger than the coefficient of x and y in that equation right it is at the third equation for that here the equation 2 becomes first equation and the second third becomes as in the second equation and equation first becomes as a third equation that is 28 x plus 4 y minus z is equal to 32 2x plus 17 y plus 4 z is equal to 35 x plus 3 y plus 10 z is equal to 24 now come to an example solve the following system of equation by jacobiz method that is 15 x plus y minus z is equal to 14 x plus 20 y plus z is equal to 23 x minus 3 y plus 18 z is equal to 37 call this is equation 1 solution given system of equation is a diagonal dominant system of equation and the and can be written for kth iteration as x suffix k is equal to 1 upon 15 into 14 minus y k minus 1 plus z k minus 1 y k is equal to 1 by 20 into 23 minus x k minus 1 minus z k minus 1 z k is equal to 1 by 18 into 37 minus x k minus 1 plus 3 into y k minus 1 call it is equation 1 let x is equal to x 0 is equal to 0 y is equal to y 0 is equal to 0 z is equal to 0 beyond the initial approximation now to find the first approximation k is equal to 1 equation 2 we get 1 upon 15 into 14 minus y 0 plus z 0 which is equal to 0.9333 y 1 is equal to 1 by 20 23 minus x 0 minus z 0 which is equal to 1.15 z 1 is equal to 1 by 18 into 37 minus x 0 plus 3 into y 0 which is equal to 2.055 second iteration put k is equal to 2 in equation 2 we get x 2 is equal to 1 by 15 into 14 minus y 1 plus z 1 which is equal to 1.0003 y 2 is equal to 1 by 20 23 minus x 1 minus z 1 which is equal to 1.0005 z 2 is equal to 1 by 18 into 37 minus x 1 plus 3 plus 3 into y 1 is equal to 2.1934 third iteration put k is equal to 3 in equation 2 we get x 3 is equal to 1 by 15 into 14 minus y 2 plus z 2 substitute the value of y 2 and z 2 from the second iteration and simplifying that is 1.0129 y 3 is equal to 1 by 20 into 23 minus x 2 minus z 2 which is equal to 0.9902 z 3 is equal to 1 by 18 into 37 minus x 2 plus 3 into y 2 which is equal to 2.1667 the fourth iteration k is equation 2 and simplifying we get x 4 is equal to 1.0117 y 4 is equal to 0.9904 z 4 is equal to 2.1643 and the fifth iteration is put the k is equal to 5 in equation 2 and simplifying the x 5 is equal to 1.0115 y 5 is equal to 0.9912 z 5 is equal to 2.1645 the fourth and fifth iteration are correct up to 3 decimal of places hence the required solution of the given equation is x is equal to 1.0115 y is equal to 0.9912 z is equal to 2.01645 references numerical solutions methods by bsgravan thank you