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Please solve using matlab code only. THank you Develop an iterative technique to solve the system of equations (1) a_11 x_1 + a_12 x_2 +
Please solve using matlab code only.
THank you
Develop an iterative technique to solve the system of equations (1) a_11 x_1 + a_12 x_2 + ... + a_1n x_n = b_1 a_21 x_1 + a_22 x_2 + ... + a_2n x_n = b_2 ... a_n1 x_1 + a_n2 x_2 + ... + a_nn x_n = b_n A vector x= vector y A =[a_11 a_21 ... a_n1 a_12 a_22 ... a_n2 ... ... ... ... a_1n a_2n ... a_nn] and vector x = [x_1 x_2 ... x_n] and vector b = [b_1 b_2 ... b_n] First rewrite the system of equations in an explicit form as: x_1 = (b_1 - x_12 x_2 - a_13 x_3 - ... - a_1n x_n)/a_11 x_2 = (b_2 - x_21 x_1 - a_23 x_3 - ... - a_2n x_n)/a_22 ... x_n = (b_n - x_n2 x_1 - a_n2 x_2 - ... - a_nn-1 x_n-1)/a_nn An initial value is assumed for each of the unknowns x^(0)_1, x^(0)_2, x^(0)_3, ... x^(0)_n. Where the number 0 in the superscript of x^(0)_n is the iteration number. For the 0^th iteration, initialize to zeros: x^(0)_i = 0, for i = 1, 2, ..., n. Substitute these values into the right hand side the of the explicit form (Eq. 2) to obtain the first approximation, x^(1)_1, x^(1)_2, x^(1)_3, ... x^(1)_n. This accomplishes one iteration. I11 the same way, the second approximation x^(2)_1, x^(2)_2, x^(2)_3, ... x^(2)_n is computed by substituting the first approximation values into the right hand side of the rewritten equations. The general form of Equations 2 is: x^k_i = 1/a_ii [Sigma^n_j=1 j notequalto i (-a_ij x^(k-1)_j) + b_i] for i = 1, 2, 3, ..., n where k is the iteration number. The k^th estimate of the solution is determined from the (k - 1)^th estimate. The iterations continue until the differences between the values that are obtained in successive iterations are small. The iterations can be stopped when the following error estimate is smaller than some tolerance | ||vector x^(k) || - || vector^(k-1)/||vector x^(k-1)||Step by Step Solution
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