Jacobi method
Example question

A linear system of the form A x = b {\displaystyle Ax=b} with initial estimate x (0) {\displaystyle x^{(0)}} is given by

We use the equation x (k + 1) = D − 1 (b − (L + U) x (k)) {\displaystyle x^{(k+1)}=D^{-1}(b-(L+U)x^{(k)})}, described above, to estimate x {\displaystyle x}. First, we rewrite the equation in a more convenient form D − 1 (b − (L + U) x (k)) = T x (k) + C {\displaystyle D^{-1}(b-(L+U)x^{(k)})=Tx^{(k)}+C}, where T = − D − 1 (L + U) {\displaystyle T=-D^{-1}(L+U)} and C = D − 1 b {\displaystyle C=D^{-1}b}. From the known values

we determine T = − D − 1 (L + U) {\displaystyle T=-D^{-1}(L+U)} as

Further, C {\displaystyle C} is found as

With T {\displaystyle T} and C {\displaystyle C} calculated, we estimate x {\displaystyle x} as x (1) = T x (0) + C {\displaystyle x^{(1)}=Tx^{(0)}+C}:

The next iteration yields

This process is repeated until convergence (i.e., until ‖ A x (n) − b ‖ {\displaystyle \|Ax^{(n)}-b\|} is small). The solution after 25 iterations is
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