Q:

Transformation of matrices in 3 dimensions

The transformation T is represented by the matrix
$$
        \begin{pmatrix}
        2 & -1 & 3 \\
        -1 & 4 & -2 \\
        3 & 2 & 4 \\
        \end{pmatrix}
$$
The plane has equation $2x-y+3z=-6$
Show that the image of the plane under T is a line and find Cartesian equations of this line.  
Let $x=s$ and $z=t$, then $y=6+2s+3t$
I know that the general point on the plane is:
$$
        \begin{pmatrix}
          s   \\
          6+2s+3t   \\
          t   \\
        \end{pmatrix}
$$
Under transformation T, it becomes:
$$
        \begin{pmatrix}
        -6 \\
        24+7s+10t \\
        12+7s+10t \\
        \end{pmatrix}
$$
$x=-6$, $y=z+12$
But I am not sure how to show that it is a line. 

A:

From here, note that $s$ and $t$ do the same thing.  $7s+10t=k$ neatly replaces all instances of $s$ and $t$ with $k$ (you get $(-6, 24+k, 12+k)$) so there is only one true degree of freedom, so it's a line.

