Linear Algebra Question, Vector Images

hemsley
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Homework Statement


Let T:R^4->R^3 be the linear transformation de fined by

T( x1, x2,x3,x4) =
2(x1) - 4(x3)
(x2) -(x3)+3(x4)
(x1)+(x2)-3(x3)+2(x4)

Find the vector from the domain, Xd, which gives the image Xr = (2 1 1) in the range of T



The Attempt at a Solution


I don't need to necessarily know the values, I just need to know the process to get the 3 space matrix to find a vector in 4 space. To go from the Vector in Xd to the image Xr, I would just need to use the numbers in the matrix transform, or just multiply the standard matrix with the vector, but I am not sure how to go the other way. Do I take the inverse of the standard matrix multiply it by Xr possibly? Please help!
 
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So you need to find (x_1,x_2,x_3,x_4) such that

T(x_1,x_2,x_3,x_4)=(2,1,1)

Or

(2x_1-4x_3,x_2-x_3+3x_4,x_1+x_2-2x_3+2x_4)=(2,1,1)

Can you make a system of equations out of this??
 
You need to use the the relation between the 4D vector (x1,x2,x3,x4) and the image vector (2,1,1):

2(x1) - 4(x3)=2
(x2) -(x3)+3(x4)=1
(x1)+(x2)-3(x3)+2(x4)=1.

This is a system of linear equations, solve it with some standard method. The matrix of the linear transformation is

T=
2 0 -4 0
0 1 1 3
1 1 -3 2

It can not be inverted.ehild

Edit: Micromass beat me ...
 
Thanks very much to both of you!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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