Solve Linear System Example (VB.NET)
[VB.NET]
' COPYRIGHT(C), SIMPLEXAR SOFTWARE LIMITED, 2006-2008.
' All rights reserved.
'
' Use of this copyright notice is precautionary only, and does not imply
' publication or disclosure. The content of this work contains confidential
' and proprietary information of Simplexar Software Limited. Any duplication,
' modification, distribution, or disclosure in any form, in whole, or in part,
' is strictly prohibited without express prior written permission.
' Get namespaces.
Imports System
Imports Simplexar.Statsar
Imports Simplexar.Statsar.LinearAlgebra
' Start namespace.
Namespace Simplexar.Examples.Statsar
' This example illustrates how to solve a linear system of equations
' by using the matrix class Solve method.
Module Example
' The application entry point.
Public Sub Main()
Try
RunExample()
Catch exception As Exception
Console.WriteLine(exception)
End Try
Console.WriteLine("Press ENTER to terminate.")
Console.ReadLine()
End Sub
Private Sub RunExample()
' To illustrate how to solve a linear system of equations,
' consider the linear system below:
'
' 2.5 * x + 3.5 * y + 4.5 * z = 45.5
' 1.5 * x + 2.8 * y + 6.2 * z = 53.48
' 8.2 * x - 1.5 * y + 3 * z = 25.14
'
' To solve this system, we first define a matrix of
' coefficients, and a vector corresponding to the right
' hand side of the linear system. We then use the matrix
' class Solve() method. This method accepts a vector and
' returns a vector. The resulting vector holds the solution
' to the linear system.
Dim m As Matrix = New Double(,) { _
{2.5, 3.5, 4.5}, _
{1.5, 2.8, 6.2}, _
{8.2, -1.5, 3}}
Dim v1 As New Vector(45.5, 53.48, 25.14)
Dim v2 As Vector = m.Solve(v1)
Dim x As Double = v2(0)
Dim y As Double = v2(1)
Dim z As Double = v2(2)
Console.WriteLine("x = {0}", x)
Console.WriteLine("y = {0}", y)
Console.WriteLine("z = {0}", z)
End Sub
End Module
End Namespace