3x3 matrices division

Thread view:

Enable EMail Alerts

Start New Thread

Thread rating:

Noa - 27 Jun 2005 20:29 GMT

Hi, University was a long time ago and I haven't used linear algebra
since...
Can someone be kind and describe how do I divide a 3x3 by another 3x3
matrix (in order to get a 1x3 vector)?
I need to write a simple code that uses this algorithm and (AFAIK)
there are no internal matrix functions in vb6.
Thx, Dan.

Reply to this Message

Randy Day - 28 Jun 2005 00:48 GMT

> Hi, University was a long time ago and I haven't used linear algebra
> since...
[quoted text clipped - 3 lines]
> there are no internal matrix functions in vb6.
> Thx, Dan.

The last example on this page touches on it:

http://www.biomechanics.psu.edu/tutorials/Kin_tutor04.html

HTH

Reply to this Message

Michael Harrington - 28 Jun 2005 06:10 GMT

I think you need to know how to invert the matrix
and also find the determinant. Here is a demo on the
latter. I'll dust off the old algebra books and see if I
can do an inversion.

Option Explicit
'Determinant of 3x3 matrix

Private Sub Command1_Click()
Dim matrix(1 To 3, 1 To 3) As Double

matrix(1, 1) = 1
matrix(1, 2) = 2
matrix(1, 3) = 0

matrix(2, 1) = -1
matrix(2, 2) = 1
matrix(2, 3) = 1

matrix(3, 1) = 1
matrix(3, 2) = 2
matrix(3, 3) = 5

Print Determinant(matrix)

End Sub

Function Determinant(M() As Double) As Double

Determinant = M(1, 1) * (M(2, 2) * M(3, 3) - M(2, 3) * M(3, 2)) _
- M(1, 2) * (M(2, 1) * M(3, 3) - M(2, 3) * M(3, 1)) _
+ M(1, 3) * (M(2, 1) * M(3, 2) - M(2, 2) * M(3, 1))

End Function

> Hi, University was a long time ago and I haven't used linear algebra
> since...
[quoted text clipped - 3 lines]
> there are no internal matrix functions in vb6.
> Thx, Dan.

Reply to this Message

spacewarp - 28 Jun 2005 19:59 GMT

Below is some code that inverses a 3 x 3 matrix. Note that the
matrices' are 1-based in my example below.
(Mind the word wraps)

Option Explicit

' This function returns the determinant of a 2 x 2 matrix
' a b
' c d
Private Function Det2D(a As Double, b As Double, c As Double, d As
Double) As Double
Det2D = a * d - b * c
End Function

' This function returns the determinant of a 3 x 3 matrix
Private Function Determinant(M() As Double) As Double
Determinant = M(1, 1) * (M(2, 2) * M(3, 3) - M(2, 3) * M(3, 2)) _
- M(1, 2) * (M(2, 1) * M(3, 3) - M(2, 3) * M(3, 1)) _
+ M(1, 3) * (M(2, 1) * M(3, 2) - M(2, 2) * M(3, 1))
End Function

' This function returns the Inverse of a 3 x 3 matrix
Private Function Inverse(M() As Double) As Double()
Dim det As Double
det = Determinant(M)

If det = 0 Then
' Handle the error as you wish ...
MsgBox "Inversion error! Matrix is singular and thus
non-invertible"
Exit Function
End If

Dim I(1 To 3, 1 To 3) As Double
I(1, 1) = Det2D(M(2, 2), M(2, 3), M(3, 2), M(3, 3)) / det
I(1, 2) = -Det2D(M(1, 2), M(1, 3), M(3, 2), M(3, 3)) / det
I(1, 3) = Det2D(M(1, 2), M(1, 3), M(2, 2), M(2, 3)) / det
I(2, 1) = -Det2D(M(2, 1), M(2, 3), M(3, 1), M(3, 3)) / det
I(2, 2) = Det2D(M(1, 1), M(1, 3), M(3, 1), M(3, 3)) / det
I(2, 3) = -Det2D(M(1, 1), M(2, 1), M(1, 3), M(2, 3)) / det
I(3, 1) = Det2D(M(2, 1), M(2, 2), M(3, 1), M(3, 2)) / det
I(3, 2) = -Det2D(M(1, 1), M(1, 2), M(3, 1), M(3, 2)) / det
I(3, 3) = Det2D(M(1, 1), M(1, 2), M(2, 1), M(2, 2)) / det

Inverse = I
End Function

' Invert a matrix when the form loads and display the results
' in the Immediate pane
Private Sub Form_Load()
Dim M(1 To 3, 1 To 3) As Double
Dim I() As Double

M(1, 1) = 1
M(1, 2) = 3
M(1, 3) = 3
M(2, 1) = 1
M(2, 2) = 4
M(2, 3) = 3
M(3, 1) = 1
M(3, 2) = 3
M(3, 3) = 4

I = Inverse(M)

Debug.Print I(1, 1), I(1, 2), I(1, 3)
Debug.Print I(2, 1), I(2, 2), I(2, 3)
Debug.Print I(3, 1), I(3, 2), I(3, 3)
End Sub

With the inversion done, it is a simple problem of multiplying the
inverse matrix with another matrix to get the result. The
multiplication part is left as an exercise to the reader ... ;-)

Hope this helps.
Suresh

Reply to this Message

VB Forum / General 2 / June 2005

3x3 matrices division