1. Linear transformation of vectors

I am in a matrix and power series class and I need to figure out how to answer the following type of question. I have read and reread the procedure the book uses and am still unable to figure this type of problem out,

I need to find all vectors in R^2 that satisfy T(v) = [-2 12 -20] or show that no such vector exists. The linear transformation T: R^2 ---> R^3 has matrix

A = [ 1 -3 1 ] T
[ 2 -8 8]

Could someone please show me the steps to attack this with? Thanks very much, Frostking

2. Originally Posted by Frostking
I am in a matrix and power series class and I need to figure out how to answer the following type of question. I have read and reread the procedure the book uses and am still unable to figure this type of problem out,

I need to find all vectors in R^2 that satisfy T(v) = [-2 12 -20] or show that no such vector exists. The linear transformation T: R^2 ---> R^3 has matrix

A = [ 1 -3 1 ] T
[ 2 -8 8]

Could someone please show me the steps to attack this with? Thanks very much, Frostking

This is a non-homogenoues linear system 3 x 2 (3 equations and 2 variables). Such a thing not always has a solution:

$\displaystyle \left(\begin{array}{cc}1&2\\ \! \! \! \! -3& \! \! \! \! -8\\1&8\end{array}\right)\left(\begin{array}{c}x\\y \end{array}\right)=\left(\begin{array}{c}-2\\\;\;12\\-20\end{array}\right) \Longleftrightarrow$ $\displaystyle \begin{array}{c}\;\;\;\;\;x+2y=\,\,\,\,\,-2\\-3x-8y=\,\,\,\,\,\,\,12\\\;\;\;\;\;\,x+8y=\;-20\end{array}$

If now you solve the first two equations as we used to do in 8th or 9th grade, we get the unique solution $\displaystyle (4,-3)$ . If you now plug in these two values in the 3rd equation you'll see you get an equality ==> there does exist a solution for the problem and it is unique (because it is unique for the first two eq's.)

Tonio
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