Let u = (4, 0) and v = (0, -3)

T is a linear transformation such that:

T(u) = (4, 4, 4)

T(v) = (6, 9, 12)

a) Find the standard matrix of T

b) Find T(x, y)

Only problem I'm having here is that I can't really see the difference between the two questions.

Here's my solution to part a:

We need to solve a(4, 0) + b(0, -3) = (x, y)

This gives us the following:

4a = x

-3b = y

a = x/4

b = -y/3

Using the linear property of transforms on a(4, 0) + b(0, -3) = (x, y), we get

T(x, y) = T[a(4, 0)] + b(0, -3)] = aT(4, 0) + bT(0, -3)

T(x, y) = a(4, 4, 4) + b(6, 9 ,12) = (x, x, x) + (-2y, -3y, -4y)

T(x, y) = (x – 2y, x – 3y, x – 4y)

which is, as far as I can see, gives me the solution to part b)!

What am I missing here?