ବ୍ୟାଖ୍ୟା 
ଗସ-ସିଡ଼ାଲ ପଦ୍ଧତି ହୋଇଛି ଏକ iterative technique ଏବଂ ଏହା ଅଜଣା x ଥିବା n ଗୋଟି linear equation ବିଶିଷ୍ଟ ଗୋଟିଏ square systemକୁ ସମାଧାନ କରିବାରେ ବ୍ୟବହାର କରାଯାଏ:
 .
iterationଦ୍ୱାରା ନିମ୍ନ ମତେ ଏହାର ସଂଜ୍ଞା ପ୍ରକରଣ କରାଯାଇପାରେ

ଏବଂ   is the kth approximation or iteration of  is the next or k + 1 iteration of , and the matrix A is decomposed into a lower triangular component , and a strictly upper triangular component U: .

ଆଲ୍‌ଗୋରିଦିମ୍ 
 Inputs: , 
 Output: 
 
 Choose an initial guess  to the solution
 repeat until convergence
     for  from 1 until  do
         
         for  from 1 until  do
             if  ≠  then
                 
             end if
         end (-loop)
         
     end (-loop)
     check if convergence is reached
 end (repeat)

ଉଦାହରଣ

ମାଟ୍ରିକ୍ସ ସଂସ୍କରଣ ପାଇଁ ଗୋଟିଏ ଉଦାହରଣ 
 ଭାବେ ଗୋଟିଏ linear system ଦର୍ଶଯାଇଛି, ଯଥା:
  ଏବଂ 
ଆମେ ନିମ୍ନୋକ୍ତ ସମୀକରଣ

କୁ ନିମ୍ନୋକ୍ତ ରୂପେ ବ୍ୟବହାର କରିବାକୁ ଚାହୁଁ

ଯେଉଁଠାରେ କି:
  ଏବଂ

ପାଇଥନ ୩ ଏବଂ NumPy ବ୍ୟବହାର କରାଯାଇଥିବା ଏକ ଉଦାହରଣ 
The following numerical procedure simply iterates to produce the solution vector.import numpy as np

ITERATION_LIMIT = 1000

# initialize the matrix
A = np.array([[10., -1., 2., 0.],
              [-1., 11., -1., 3.],
              [2., -1., 10., -1.],
              [0.0, 3., -1., 8.]])
# initialize the RHS vector
b = np.array([6., 25., -11., 15.])

# prints the system
print("System:")
for i in range(A.shape[0]):
    row = ["{}*x{}".format(A[i, j], j + 1) for j in range(A.shape[1])]
    print(" + ".join(row), "=", b[i])
print()

x = np.zeros_like(b)
for it_count in range(ITERATION_LIMIT):
    print("Current solution:", x)
    x_new = np.zeros_like(x)

    for i in range(A.shape[0]):
        s1 = np.dot(A[i, :i], x_new[:i])
        s2 = np.dot(A[i, i + 1:], x[i + 1:])
        x_new[i] = (b[i] - s1 - s2) / A[i, i]

    if np.allclose(x, x_new, rtol=1e-8):
        break

    x = x_new

print("Solution:")
print(x)
error = np.dot(A, x) - b
print("Error:")
print(error)ଏହାର output ଆସିବ:System:
10.0*x1 + -1.0*x2 + 2.0*x3 + 0.0*x4 = 6.0
-1.0*x1 + 11.0*x2 + -1.0*x3 + 3.0*x4 = 25.0
2.0*x1 + -1.0*x2 + 10.0*x3 + -1.0*x4 = -11.0
0.0*x1 + 3.0*x2 + -1.0*x3 + 8.0*x4 = 15.0

Current solution: [ 0.  0.  0.  0.]
Current solution: [ 0.6         2.32727273 -0.98727273  0.87886364]
Current solution: [ 1.03018182  2.03693802 -1.0144562   0.98434122]
Current solution: [ 1.00658504  2.00355502 -1.00252738  0.99835095]
Current solution: [ 1.00086098  2.00029825 -1.00030728  0.99984975]
Current solution: [ 1.00009128  2.00002134 -1.00003115  0.9999881 ]
Current solution: [ 1.00000836  2.00000117 -1.00000275  0.99999922]
Current solution: [ 1.00000067  2.00000002 -1.00000021  0.99999996]
Current solution: [ 1.00000004  1.99999999 -1.00000001  1.        ]
Current solution: [ 1.  2. -1.  1.]
Solution:
[ 1.  2. -1.  1.]
Error:
[  2.06480930e-08  -1.25551054e-08   3.61417563e-11   0.00000000e+00]

Program to solve arbitrary no. of equations using Matlab 
The following code uses the formula
function [x] = gauss_seidel(A, b, x0, iters)
    n = length(A);
    x = x0;
    for k = 1:iters
        for i = 1:n
            x(i) = (1/A(i, i))*(b(i) - A(i, 1:n)*x + A(i, i)*x(i));
        end
    end
end

ଆହୁରି ଦେଖନ୍ତୁ 
 Jacobi method
 Successive over-relaxation
 Iterative method. Linear systems
 Gaussian belief propagation
 Matrix splitting
 Richardson iteration

ଟୀକା

ଆଧାର 
 .
 .

ବାହାର ଲିଙ୍କ 
 
 Gauss–Seidel from www.math-linux.com
 Module for Gauss–Seidel Iteration
 Gauss–Seidel From Holistic Numerical Methods Institute
 Gauss Siedel Iteration from www.geocities.com
 The Gauss-Seidel Method
 Bickson
 Matlab code 
 C code example

ବିଜ୍ଞାନ