SOLVED: Computing a vector without knowing A

[Ricemarine]

Edit: I checked the super long way of plugging in, and yeah, those are the answers.

So I have 3 vectors [1,1,1] [2,1,2], and [5,5,6]. When transformed (multiplied, something...) with A they become [2,3,4] [4,3,2], and [-2,-3,-4].

So now the question is, what is A of [7,3,0]

So what I tried doing is doing an augmented matrix with the original 3 vectors and on the augmented side putting [7,3,0]

[1 2 5 | 7]
[1 1 5 | 3]
[1 2 6 | 0]

that then reduces to x1 = 34, x2 = 4, x3 = -7.

So then I use those weights and multiply that by the transformed vectors.

[ 2 4 -2 ] [ 34 ]
[ 3 3 3 ] [ 4 ]
[ 4 2 -4 ] [-7 ]

becomes

[ 98 ]
[ 135 ]
[ 172 ]

But I'm wondering if that's right because that seems to be a high number compared to the other three vectors that were multiplied by A.


Thanks...

 

[Ricemarine]

Ok who's the one that keeps rating all the threads 0?

 

[ItTheCow]

You start with a 3x3 matrix ( [1,1,1] [2,1,2], and [5,5,6] ), and multiply it with A to get another 3x3 matrix ( [2,3,4] [4,3,2], and [-2,-3,-4] ). That means A is either a scalar, or another 3x3 matrix. The solution you posted, (x1 = 34, x2 = 4, x3 = -7) is a 3x1, and cannot be correct.

 

[Kelemvor]

Originally posted by: Ricemarine
Ok who's the one that keeps rating all the threads 0?

I'll give you an Excellent. heh heh.

 

[oboeguy]

The input vectors make-up a 3x3 matrix. The outputs too. Call them B and C. You have a set of equations now, A*B=C. Lucky you, B is invertible so you can compute A explicitly. Get to work! (or fire-up Matlab FTW)