Points and Matrices Question:
Consider in $\mathbb{R}^2$, the points satisfying the equation $2x_2-x_1-2 = 0$. Show on a plot the points satisfying this equation. How are these points transformed by the matrix
$$
A =
\begin{pmatrix}
2 & 1 \\
1 & 1
\end{pmatrix}
$$
I understand that I'm given the equation of a line. But I don't understand how to plot all points that satisfy the equation - isn't it infinite? also, I don't understand how the points are affected by the matrix. Are they just scaled?
 A: The equation $2x_2-x_1-2=0$ is that of an affine line $L$ in $\mathbb{R}^2$. To plot this, you can for instance put it in the form
$$
y=\frac{1}{2}x+1
$$
which shows that the slope is $1/2$ and that the line passes through $(0,1)$.
Since $A$ is an invertible linear transformation, it transforms the latter line into another line. To get the equation of this new line, compute first
$$
A\left( \matrix{x\\y}\right)=\left(\matrix{z=2x+y\\t=x+y}\right).
$$
Now if $y=\frac{1}{2}x+1$, we get $z=\frac{5}{2}x+1$ and $t=\frac{3}{2}x+1$, so 
$$
5t-3z+2=0.
$$
So the image of a point on the line $L$ belongs to the line of equation
$$
5t-3z+2=0\quad\Leftrightarrow\quad t=\frac{3}{5}z-\frac{2}{5} 
$$
where $t$ is the ordinate, and $z$ the abscissa.
Now observe that you can move backwards and show conversely that every point on the latter is in the image of $L$, ie given $(z,t)$ satisfying the above, we can find $(x,y)$ on $L$ such that $A(x,y)=(z,t)$.
So the image of $L$ under $A$ is the line $L'$ with the equation given above.
