Page 1 :
systems of equations yield solution after an, computation that can be specified in advance. But in iterative, methods (or indirect methods) we first develop a rule to find the, SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS, Gauss-Jordan methods are called direct methods since the given, 1.93, O ITERATIVE METHODS, The previous two methods viz., Gaussian elimination,, amount of, hest possible solution. We start with an initial approximate, solution and apply the rule to get a better solution. This solution is, again subjected to the rule to get a still better solution and so on. If, the rule is applied repetitively, then each successive calculation to, determine the next approximation to the solution is called an, iteration. The successive approximations themselves are called, iterates., Note : Iteration method is self-correcting method, since, the error made in any computation is corrected in the, subsequent iterations., O Gauss - Seldel Iterative Method, Let the given system of equations be, a11*1 + a12X2+ a13X3+, + a\n*n, %3D, ......, a21*1 + a22x2 + a23*3 +, + a2n*n, C2, ......, a31x1 + a32x2 + a33X3 +, + azn*n =, ......, anj*1 + an2*2+ an3*3 + ......, + ann*n, Such system is often amenable to an iterative process in which, the system is first rewritten in the form, 1., au (C - a12x2–- a13*3-, .. (1), X1, a11, (C1 - a21*1 – a23x3 – ... - a2Xn), a22, · (2), X2 =, X3 =, a33, (C3 - az1x1 - d32x2-, ... (3), ......, - an, n– 1 *n - 1) .. (4), %3D, ann, UNIT 1, Scanned by CamScanner

Page 2 :
we can find the better, = Xn = 0 in (1) and, (2) we get the value for x2 and let it be x2*. Putting x* for x1 and, 0 in (3) we get the value for x3, NUMERICAL METHODS, 1.94, First let us assume that x2 = x3, Putting x, %3D, ..... ...., * for x1 and x3 = x4 =, %3!, find xj. Let it be x*., %3D, %3D, for x and x3 = X4 = ..., approximate, Xp. Similarly, x, by using the relation, ......., values for x1, X2,, ......,, approximate value of x, x2,, x1* * a11, (C - a12*2- 413*3 ., x2* a22, (C2 – az1* * – A23*3-, %3D, (C3 - az1x1* – a32*2*, a33, (C,-an*1* - An2*2* - An3X3*, ann, - ап, п-1 *п-1), :, large number of zero elements., Note 2: We say a matrix is diagonally dominant if 4., numerical value of the leading diagonal element in each row, greater than or equal to the sum of the numerical values of a, other elements in that row., 5 1 -1, 1 4, 2 is diagonally, For example the matrix, 1 -2, 5., dominant., 5 1 -1, 5 2, 1 -2, row, the leading diagonal element 2 is less than the sum 8 of, the other two elements viz., 5 and 3 in that row., Note 3 : For the Gauss - Seidal method to converge quickly,, the coefficient matrix must be diagonally dominant. If it is not, 50, we have to rearrange the equations in such a way that the, coefficient matrix is diagonally dominant and then only we cau, 3 is not, since in the second, 5, But the matrix, apply Gauss-Seidal method., I UNIT 1, Scanned by CamScanner

Page 3 :
SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS, 1.95, 8 -3 2, E.g. 1: The coefficient matrix, 3 12, 4 11 -1, can be made, diagonally dominant by interchanging second and third rows, 8 -3 2, 4 11 -1, as, 3 12, E.g. 2: Consider the system of equations, x1 + 7x2 - x3, = 3, 5x1 + x2 + X3, = 9, -3x1 + 2x2 + 7x3, = 17, 1 7 -1, 5 1, The coefficient matrix, is not diagonally, 7, 1, -3 2, dominant., But the given system can be rearranged like, 7x2 + X1 - X3, 3, %3D, X2 + 5x1 + x3, 2x2 - 3x1 + 7x3, = 9, = 17, 7 1 -1, 1 5, 2 -3, When solved the solution appear in the order x2, x1 and x3., Now the coefficient matrix, 1 is diagonally dominant., 6 Example 1 d, Solve by Gauss – Seidel method, X- 2y, 2x + 25y, - 3, = 15, [A.U. May 2000], Solution, The given system is, x- 2y = - 3, 2x + 25y = 15, ... (1), .. (2), From (1) and (2) we get,, x = -3+ 2y, 1, (3), y =, 25, [15-2r], . (4), UNIT 1 E, Scanned by CamScanner

Page 4 :
NUMERICAL METHODS, 1.96, First Iteration:, Putting y = 0 in (3) we get, x = - 3, Putting x = -3 in (4) we get, %3D, 0-84, [15 – 2 (– 3)], y =, 25, : In the first iteration we get, 3, y = 0-84, Second Iteration, Putting y = 0-84 in (3) we get, x = - 3 + 2 (0-84)], - 1:32, %3D, Putting x = - 1-32 in (4) we get, 자[15-2 (-1·32)] = 0-7056, %3D, アミ, 25, .: In the second iteration we get, x = - 1-32 , y = 0-7056, Third Iteration, Putting y = 0-7056 in (3) we get, x = - 3+2 (0-7056)] = - 1-589, Putting x =- 1:589 in (4) we get, %3D, 25 (15 – 2 (- 1-589)] = 0-727, y =, %3D, :. In the third iteration we get, x = - 1:589 ,, y = 0-727, Fourth Iteration, Putting y = 0-727 in (3) we get, x = -3+ 2 (0-727)] = – 1-546, %3D, Putting x = - 1:546 in (4) we get, y =, [15 – 2 (- 1-546)] = 0-724, 25, :. In the fourth iteration we get, * = - 1:546 , y = 0-724, UNIT 1, Scanned by CamScanner

Page 5 :
SOLUTION OF EQUATIONS AND EIGENVALUE PROBLEMS, 1.97, Fifth Iteration, Putting y = 0-724 in (3) we get, %3D, x = - 3 + 2 (0-724)], - 1-552, Putting x = - 1:552 in (4) we get, y =, 25, [15 – 2 (- 1:552)], = 0:724, :: In the fifth iteration we get, = x, 1-552 , y = 0-724, Sixth Iteration, Putting y = 0-724 in (3) we get, x = - 3 + 2 (0-724)], = - 1:552, Putting x = - 1-552 in (4) we get, y =, 5 (15 – 2 (- 1-552)], 25, = 0-724, :: After sixth iteration we get, = x, 1-552 , y = 0-724, %3D, Checking :, х- 2у, - 1:552 – 2 (0-724) =, * Example 2 o, Solve the system of equations 4x + 2y +z = 14, x + 5y – z = 10,, x+y+8z = 20 using Gauss - Seidel iteration method., - 3, %3D, - 3, %3D, %3D, Solution, The given system is, 4x + 2y + z = 14, x + 5y -z = 10, x+ y+ 8z = 20, .. (1), -(2), .. (3)., %3D, Clearly the coefficient matrix, 1, 4 2 1, 15 -1, 1 8, is diagonally, dominant. Hence we can apply Gauss - Seidel method without any, difficulty., UNIT 1, Scanned by CamScanner