Originally Posted by

**Deveno** my totally naive approach:

the formula for T (provided we CAN find one) is going to be something like:

T(x,y) = (ax+by,cx+dy) <---general linear function from R^{2} to R^{2}.

so what we really want to do is find a,b,c and d.

now T(4,7) = (3,4). to me this means:

4a+7b = 3

4c+7d = 4

that by itself isn't enough to say what a,b,c and d are, but it's a start.

we also know T(3,5) = (4,9) so:

3a+5b = 4

3c+5d = 9

looking at just a&b, we have the two equations:

4a+7b = 3

3a+5b = 4

so:

12a+21b = 9

-12a-20b = -16

thus b = -7 (and then from 4a+7b = 3, we get: 4a - 49 = 3, so 4a = 52, thus a = 13).

similarly:

12c+21d = 12

=12c-20d = -36

so d = -24 (and then from 4c+7d = 4, we get: 4c - 168 = 4, so 4c = 172, so c = 43).

so our formula for T is:

T(x,y) = (13x-7y,43x-24y)

yes, you CAN consider a basis B, and then find the matrix of T relative to that basis. but since we're in R^{2}, and using coordinates (x,y) to denote elements of R^{2}, we can by-pass that altogether, and just deal with "coordinate functions":

T_{1}(x,y) = ax+by (the first coordinate of T(x,y))

T_{2}(x,y) = cx+dy (the second coordinate of T(x,y)).

again, for emphasis: linear functions are linear in EACH coordinate.

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NOTE: there is nothing wrong with any of the previous answers. realize that if you are going to find a FORMULA for a linear transformation T in terms of COORDINATES, one really should say "which basis the coordinates are IN". writing (x,y) usually tacitly assumes that the basis is B = {(1,0),(0,1)}, but the matrix for T will have "different numbers" (i.e., a different formula), if you use some OTHER basis (in the domain, range, or both).