Find image under $T$ of the line $x_1 + 2x_2 = 3$ -- Linear Algebra We are asked to find the image under $T$ of the line $$x_1 + 2x_2 = 3$$.
Consider the linear operator $T:\mathbb{R^2}\rightarrow \mathbb{R^2}$ with standard matrix $$ A = \left[\begin{array}{rrr}
    1 & 1 \\
    0 & 1 \\
    \end{array}\right]
$$
The line $x_1 + 2x_2 = 3$ contains the points $$(-1,2), (1,1), (3,0)$$
I do not understand how they arrive at the aforementioned points.
 A: 
I do not understand how they arrive at the aforementioned points.

Make $x_1 = -1$. Solve for $x_2$ and get $x_2 = 2$. So $(-1,2)$ is in the line.
Make $x_1 =1$. Solve for $x_2$ and get $x_2 = 1$. So $(1,1)$ is in the line.
Make $x_1 = 3$. Solve for $x_2$ and get $x_2 = 0$. So $(3,0)$ is in the line.
We can't help you further since we don't know that $T$ is. Now we know. You can write: $$[T(x_1,x_2)] = \begin{bmatrix} 1 & 1 \\ 0 & 1\end{bmatrix}\begin{bmatrix} x_1 \\ x_2\end{bmatrix} = \begin{bmatrix} x_1+x_2 \\ x_2\end{bmatrix}$$Since $T$ is invertible, $T$ will take that line to another line. And you only need two points, instead of three. Compute $T(-1,2)$, $T(1,1)$, and the line passing through these points. Now it's on you!
A: you can parametrise the line by $$(x_1, x_2)^\top =(-1,2)^\top  + t(2, -1)^\top. $$ the point $(-1,2)^\top$ is mapped by $T$ to $-1(1,0)^\top+2(1,1)^\top = (1, 2)^\top$ and $(2,-1)^\top$ is mapped by $T$ to $2(1,0)^\top -(1,1)^\top = (1, -1)^\top$ 
therefore $$A(x_1,x_2)^\top =(1,2)^\top+t(1,-1)^\top. \tag 1  $$
$(1)$  represents the line $$x_1 + x_2 = 3. $$
