Take $x$ to be the vector $[x,y,z]^T$ and multiply it by your matrix $A$. That would be your linear transformation.
The 'image' of a vector under a function just means the value of the function when that vector is put in as an argument. So do $T(v)$ and $T(u)$ once you find out $T$ explicitly.
Well, first of all you should know that given a linear transformation $T : V \to W$ if you know the values of $T$ on the basis of $V$ then you know $T$ completely by virtue of linearity. The point here is: in an $n$-dimensional vector space, $n$ linearly independent vectors necessarily forms a basis.
In this case $\mathbb{R}^2$ has dimension $2$ so that any set of $2$ linearly independent vector is a basis for $\mathbb{R}^2$. It's easy to see that $v_1$ and $v_2$ are linearly independent because they're not multiple one of the other. Hence the set $\left\{v_1, v_2\right\}$ is a basis of $\mathbb{R}^2$.
Now here comes the point. Given any $(x,y) \in \mathbb{R}^2$, how do we write it in this new basis? Well, it's easy, it should be a linear combination, so that we must have:
$$(x,y) = av_1+bv_2$$
And substituting we get:
$$(x,y) = a(1, -1) + b(2, -3)$$
$$(x,y) = (a+2b, -a-3b)$$
This is a system of linear equations in $a$ and $b$. If you solve, you'll find that $a = 3x+2y$ and $b =-x-y$. So that any vector $(x,y) \in \mathbb{R}^2$ is given in that basis by:
$$(x,y) = (3x+2y)v_1 + (-x-y)v_2$$
And hence
$$T(x,y) = (3x+2y)T(v_1) + (-x-y)T(v_2)$$
Now it's just a question of substituting the values:
$$T(x,y) = (3x+2y)(7, -8) + (-x-y)(17,-19)$$
So that finally, the expression of $T$ on arbitrary $(x,y)\in \mathbb{R}^2$ is:
$$T(x,y) = (4x-3y, -5x+3y)$$
Which is really the expression of $T$. You can easily verify that $T(v_1)$ and $T(v_2)$ are what they are expected to be.
Best Answer
Remember that $T$ is linear. That means that for any vectors $v,w\in\mathbb{R}^2$ and any scalars $a,b\in\mathbb{R}$, $$T(av+bw)=aT(v)+bT(w).$$ So, let's use this information. Since $$T \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 12 \\ -2 \end{bmatrix}, \qquad T\begin{bmatrix} 2 \\ -1 \\ \end{bmatrix} = \begin{bmatrix} 10 \\ -1 \\ 1 \end{bmatrix},$$ you know that $$T\left(\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}+2\begin{bmatrix} 2 \\ -1 \\ \end{bmatrix}\right)=T\left(\begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}+\begin{bmatrix} 4 \\ -2 \\ \end{bmatrix}\right)=T\begin{bmatrix} 5 \\ 0\\ \end{bmatrix}$$ must equal $$T \begin{bmatrix} 1 \\ 2 \\ \end{bmatrix}+2\cdot T\begin{bmatrix} 2 \\ -1 \\ \end{bmatrix} =\begin{bmatrix} 0 \\ 12 \\ -2 \end{bmatrix}+2\cdot \begin{bmatrix} 10 \\ -1 \\ 1 \end{bmatrix}=\begin{bmatrix} 20 \\ 10 \\ 0 \end{bmatrix}.$$ So, we know $T\begin{bmatrix} 5 \\ 0\\ \end{bmatrix}$. Do you see how to find $T\begin{bmatrix} 1 \\ 0\\ \end{bmatrix}$? Then use the same process to figure out $T\begin{bmatrix} 0 \\ 1\\ \end{bmatrix}$.
After doing that, you should know how to make the (standard basis) matrix for $T$.