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
so we want a vector $(x,y,z)$ such that when premultiplied by A gives us b.
so therefore $x+y-2z=3$, $y+z=-1 $and $-2x-y-5z=-1$. We are allowed multiply these equations by constants and add these equations. (left side with left and right with right. As you can see, (if we ignore the variables and just look at the coefficients this is the same as adding one row of the matric to another and multiplying one row with another. Therefore we can work with matrices. Where each row contains the coefficients of the variables on the first three columns and the constant to which they add up on the right side.
So at the start we have.
$$\begin{bmatrix}1&1&-2&3\\0&1&1&-1\\-2&-1&5&-7\end{bmatrix}$$
Then we switch the second and third rows
$$\begin{bmatrix}1&1&-2&3\\-2&-1&5&-7\\0&1&1&-1\end{bmatrix}$$
add twice the first one to the second one
$$\begin{bmatrix}1&1&-2&3\\0&1&1&-1\\0&1&1&-1\end{bmatrix}$$
and then subtract the second one from the third one to get
$$\begin{bmatrix}1&1&-2&3\\0&1&1&-1\\0&0&0&0\end{bmatrix}\;.$$
We can simplify a bit more by subtracting the second row from the first:
$$\begin{bmatrix}1&0&-3&4\\0&1&1&-1\\0&0&0&0\end{bmatrix}\;.$$
This says that $x_1-3x_3=4$ and $x_2+x_3=-1$, or
$$\left\{\begin{align*} x_1&=4+3x_3\\ x_2&=-1-x_3\;, \end{align*}\right.$$
where $x_3$ can be any real number whatsoever. Taking $x_3=0$ makes the calculation very easy and gives us one solution:
$$x=\begin{bmatrix}4\\-1\\0\end{bmatrix}\;.$$
But this is clearly not unique: we get infinitely many other solutions by changing the value of $x_3$.