initialapprox = Vector or Matrix

The initial approximation. To obtain a float solution instead of an exact solution, the initialapprox should contain floats instead of integers. By default, a zero vector is used.

maxiterations = posint

The maximum number of iterations to perform while approximating the solution to $A·x=b$.  If the maximum number of iterations is reached and the solution is not within the specified tolerance, a plot of distances can still be returned. By default, maxiterations = 20.

method = jacobi, gaussseidel, SOR(numeric), LU, LU[tridiagonal], PLU, or PLU[scaled]

The method to use when solving the linear system $A·x=b$. Please see the Notes section for sufficient conditions for convergence. This option is required. Each method is described below:

jacobi : The Jacobi method. Optionally, the stoppingcriterion, maxiterations, initialapprox and tolerance options may be specified as well.

gaussseidel : The Gauss-Seidel method. Optionally, the stoppingcriterion, maxiterations, initialapprox and tolerance options may be specified as well.

SOR(w) : The Successive Overrelaxation method with w as its extrapolation factor. Optionally, the stoppingcriterion, maxiterations, initialapprox and tolerance options may be specified as well.

LU and LU[tridiagonal] : LU Decomposition. This method performs LU factorization on A and then solves the subsequent systems. None of the remaining options are used with this method. An error will be raised if the LU[tridiagonal] method is specified and A is not tridiagonal.

PLU and PLU[scaled] :  PLU Decomposition. This method performs PLU factorization on A and then solves the subsequent systems. None of the remaining options are used with this method.

stoppingcriterion = distance(norm)

The stopping criterion for an iterative technique; it is of the form stoppingcriterion=distance(norm), where distance is either relative or absolute and norm is a nonnegative integer, infinity, or Euclidean. By default, stoppingcriterion=relative(infinity).

tolerance = positive

The tolerance of the approximation. By default, a tolerance of $\frac{1}{10000}$ is used.

The LinearSolve command numerically approximates the solution to the linear system $A·x=b$, using the specified method.

The IterativeApproximate command and the MatrixDecomposition command are both used by the LinearSolve command.

If b is a matrix, then the systems $A·x=b_i$ will be solved for each column $b_i$ of $b$, and hence there will be multiple solutions returned.

Different options are required to be specified in opts, depending on the method.  These dependencies are outlined in the Options section.

The Notes section in the Student[NumericalAnalysis][IterativeApproximate] help page lists conditions under which the Jacobi, Gauss-Seidel, and successive over-relaxation iterative methods produce a solution.

$\mathrm{with}\left(\mathrm{Student}\left[\mathrm{NumericalAnalysis}\right]\right)\:$

$A≔\mathrm{Matrix}\left(\left[\left[10.\,-1.\,2.\,0.\right]\,\left[-1.\,11.\,-1.\,3.\right]\,\left[2.\,-1.\,10.\,-1.\right]\,\left[0.\,3.\,-1.\,8.\right]\right]\right)$

$A≔\left[\begin{array}{cccc}10.& \mathrm{-1.}& 2.& 0.\\ \mathrm{-1.}& 11.& \mathrm{-1.}& 3.\\ 2.& \mathrm{-1.}& 10.& \mathrm{-1.}\\ 0.& 3.& \mathrm{-1.}& 8.\end{array}\right]$

$b≔\mathrm{Vector}\left(\left[6.\,25.\,-11.\,15.\right]\right)$

$b≔\left[\begin{array}{c}6.\\ 25.\\ \mathrm{-11.}\\ 15.\end{array}\right]$

$\mathrm{LinearSolve}\left(A\,b\,\mathrm{method}=\mathrm{SOR}\left(1.25\right)\,\mathrm{initialapprox}=\mathrm{Vector}\left(\left[0.\,0.\,0.\,0.\right]\right)\,\mathrm{maxiterations}=100\,\mathrm{tolerance}={10}^{-4}\right)$

$\left[\begin{array}{c}0.9999776440\\ 2.000001578\\ \mathrm{-0.9999942334}\\ 0.9999867498\end{array}\right]$

$\mathrm{LinearSolve}\left(A\,b\,\mathrm{method}=\mathrm{LU}\right)$

$\left[\begin{array}{c}1.000000000\\ 2.000000000\\ \mathrm{-1.000000000}\\ 0.9999999999\end{array}\right]$

$B≔\mathrm{Matrix}\left(\left[\left[6.\,25.\,-11.\,15.\right]\,\left[7.\,8.\,16.\,4.\right]\,\left[4.\,2.\,9.\,5.\right]\,\left[17.\,6.\,3.\,22.\right]\right]\right)$

$B≔\left[\begin{array}{cccc}6.& 25.& \mathrm{-11.}& 15.\\ 7.& 8.& 16.& 4.\\ 4.& 2.& 9.& 5.\\ 17.& 6.& 3.& 22.\end{array}\right]$

$\mathrm{LinearSolve}\left(A\,B\,\mathrm{method}=\mathrm{PLU}\right)$

$\left[\left[\begin{array}{c}0.5095334686\\ 0.1482082486\\ 0.5264367816\\ 2.135226505\end{array}\right]\,\left[\begin{array}{c}2.623529412\\ 0.8352941177\\ \mathrm{-0.2000000000}\\ 0.4117647058\end{array}\right]\,\left[\begin{array}{c}\mathrm{-1.210953347}\\ 1.465179175\\ 1.287356321\\ \mathrm{-0.01352265042}\end{array}\right]\,\left[\begin{array}{c}1.376064909\\ \mathrm{-0.2600405681}\\ 0.4896551724\\ 2.908722110\end{array}\right]\right]$