# Tranforming a matrix

Posted by Anonymous User
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 Anonymous User Tranforming a matrix September 10, 2006 02:01PM IP/Host: ---.saix.net Rating: 0
[-1.40787]
[1.49657 ]
[-3.12701]
[1.000000]

Transforming the matrix D with T
Translation Matrix T:
[ 1.0...0.00...0.00...4.0]
[0.00...1.0....0.00...5.0]
[0.00...0.00...1.0....2.0]
[0.00...0.00...0.00...1.0]
Matrix Multiply TxD:
[( 1.0*-1.4 )+( 0.00*1.5 )+( 0.00*-3.1 )+( 4.0*1.0 )
[( 0.00*-1.4 )+( 1.0*1.5 )+( 0.00*-3.1 )+( 5.0*1.0 )]
[( 0.00*-1.4 )+( 0.00*1.5 )+( 1.0*-3.1 )+( 2.0*1.0 )]
[( 0.00*-1.4 )+( 0.00*1.5 )+( 0.00*-3.1 )+( 1.0*1.0 )]
Answer : (new location of point)
[2.59213]
[6.49657]
[-1.12701]
[1.00000]

Does this look right?
 Anonymous User Re: Tranforming a matrix September 21, 2006 01:08PM IP/Host: ---.saix.net Rating: 0
Hi

I get the same result, so your calculation seems to be correct. I assume that you are using 4.0, 5.0, 2.0 as your displacement vector in the translation.

If we use the same values for the sake of computation, then for scaling we would get:

[-1.40787]
[1.49657 ]
[-3.12701]
[1.000000]

Scaling Matrix S

[4.0...0.00...0.00...0.0]
[0.00...5.0....0.00...0.0]
[0.00...0.00...2.0....0.0]
[0.00...0.00...0.00..1.0]

and then

[-5.63148]
[7.48285]
[-6.25402]
[1.00000]

Do you agree?
 Anonymous User Re: Tranforming a matrix September 21, 2006 04:55PM IP/Host: ---.saix.net Rating: 0
Yes I get the same answer

[-5.63148]
[7.48285]
[-6.25402]
[1.00000]

but

This is assuming we are using the 3 dimensional homogenous point (4.0, 5.0, 2.0, 1) on the coordinates basis vectors (1,0,0) , (0,1,0) and (0,0,1)

 Anonymous User Re: Tranforming a matrix October 02, 2006 10:28AM IP/Host: ---.saix.net Rating: 0
Im not following you here!
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