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In-Place Matrix Inversion by Modified Gauss-Jordan Algorithm

D. DasGupta, M.Tech., M.ASCE, P.E., MCP

Former V.P.-Development, LEAP Software, Inc., Tampa, FL, USA
Former Principal Consultant, McDonnell Douglas Automation Co., St. Louis, MO, USA
Former Assistant Director, Central Water and Power Commission, New Delhi, India

REF: Applied Mathematics, 2013, 4, 1392-1396 http://dx.doi.org/10.4236/am.2013.410188 Published Online October 2013

     The classical Gauss-Jordan method for matrix inversion involves augmenting the matrix with a unit matrix and requires a workspace twice as large as the original matrix as well as computational operations to be performed on oth the original and the unit matrix. A modified version of the method for performing the inversion without explicitly generating the unit matrix by replicating its functionality within the original matrix space for more efficient utilization of computational resources is presented in this article.
     Although the algorithm described here picks the pivots solely from the diagonal which, therefore, may not contain a zero, it did not pose any problem for the author because he used it to invert structural stiffness matrices which met this requirement. Techniques such as row/column swapping to handle off-diagonal pivots are also applicable to this method but are beyond the scope of this article.

Keywords: Numerical Methods, Gauss-Jordan, Matrices, Inversion, In-Place, In-Core, Structural Analysis

Excerpt of the document:

Modified Gauss Jordan algorithm
classical matrix inversion 2
explanation of inversion 3
example of inversion

verification of inversion

in place, modified method for inversion

step 1 modified version
step 2
inversion complete after 3 iterations


Microsoft® Visual Basic® Routine for In-Place Matrix Inversion

'(A) Sample calling routine

'Sample matrix size
 Dim iNmat As Integer
 iNmat = 3 

'Sample matrix elements
 ReDim dblMatrix(iNmat, iNmat) As Double
 dblMatrix(1, 1) = -1: dblMatrix(1, 2) = 1: dblMatrix(1, 3) = 1
 dblMatrix(2, 1) = 1.2: dblMatrix(2, 2) = -1: dblMatrix(2, 3) = -1.6
 dblMatrix(3, 1) = 0.4: dblMatrix(3, 2) = 0: dblMatrix(3, 3) = -0.2 

'In-Place Inversion
 InPlace iNmat, dblMatrix() 

'The matrix dblMatrix will now contain the inverse


'(B) In-Place Matrix Inversion Subroutine

 Public Sub InPlace(iNmat As Integer, arrMat() As Double) 

'iNmat       = Matrix size
'arrMat      = Matrix array 

 Dim iRow As Integer, jCol As Integer, iCycle As Integer
 Dim dPivot As Double, dAbsDiag As Double, iBig As Integer
 ReDim blnRowProcessed(iNmat) As Boolean 

'Preset entire Index vector to Active
 For iRow = 1 To iNmat: blnRowProcessed(iRow) = False: Next iRow 

'Loop over degrees of freedom
 For iCycle = 1 To iNmat 

'  Preset Pivot to Zero
   dPivot = 0! 

'  Find the row number and value of largest diagonal element of
'  active rows (Index = 1) 

'  Loop over matrix rows
   For iRow = 1 To iNmat
'    If Index for this row is YES
     If blnRowProcessed(iRow) = False Then
'       Store ABS value of its diagonal element
        dAbsDiag = Abs(arrMatrix(iRow, iRow))
'       If ABS(Diagonal) > Current Pivot then
        If dAbsDiag > dPivot Then
'          Store current row number and diagonal value as pivot
'          Note:  Since all diagonal elements must be positive,
'          this test must succeed at least once
           iBig = iRow: dPivot = dAbsDiag
        End If
      End If

'   The number of active row with highest diagonal has been determined
'   and the value of its diagonal element has been stored as Pivot 

'   Reset Index of the row with highest diagonal to Inactive and its
'   diagonal element to Unity.
    blnRowProcessed(iBig) = True
    dPivot = arrMatrix(iBig, iBig)
    arrMatrix(iBig, iBig) = 1! 

'   Divide the entire Pivotal row with Pivot
    For jCol = 1 To iNmat
      arrMatrix(iBig, jCol) = arrMatrix(iBig, jCol) / dPivot

'   This will make the Pivotal Row diagonal = unity
'   the reciprocal of its original value? 

'   Loop over all rows
    For iRow = 1 To iNmat   

'     Skip the Pivotal row which has already been processed
      If iRow <> iBig Then       

'        Retrieve the Pivotal column element of the current row and
'        reset it to zero
         dPivot = arrMatrix(iRow, iBig)
         arrMatrix(iRow, iBig) = 0!           

'        Subtract from each element of the current row the
'          Pivotal column element times the Pivotal column element in
'          the row equal to current column 

         For jCol = 1 To iNmat
           arrMatrix(iRow, jCol) = arrMatrix(iRow, jCol) - _
             dPivot * arrMatrix(iBig, jCol)

      End If       



End Sub

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