Be $M$ the linear transformation represented by matrix $T$:
$$T =
\left[ \begin{matrix}
5 & 2 \\
4 & 1 \\
\end{matrix}\right ]
$$
Mark the correct answer which show the image by $M$ of the line $2x-3y = 0$
(A) $19x – 8y = 0$
(B) $19x – 14y = 0$ (answer)
(C) $19x -15y = 0$
(D) $15x-14y = 0$
(E) $4x – 5y = 0$
Any hints?
My first attempt was related the tranformation the each coordanaties from line r: 2x-3y = 0 with matrix T.
r as a vector: $\vec{r} = \left[\begin{matrix} 2\\-3\end{matrix}\right]$
$\vec{z} = T . \vec{r} = \left[\begin{matrix} 4\\-5\end{matrix}\right]$ (wrong)
Best Answer
The simplest way to obtain the image of the given line
$$2x-3y = 0$$
is to obtain images of its $2$ points - why don't select the natural points $A=[0,0],\ B=[3,2]$?
Computing their images we obtain
$$\left( \begin{matrix} 5 & 2 \\ 4 & 1 \\ \end{matrix}\right )\cdot \left( \begin{matrix} 0 \\ 0 \\ \end{matrix}\right ) = \color{red}{ \left( \begin{matrix} 0\\ 0 \\ \end{matrix}\right )},\quad\left( \begin{matrix} 5 & 2 \\ 4 & 1 \\ \end{matrix}\right )\cdot \left( \begin{matrix} 3 \\ 2 \\ \end{matrix}\right ) = \color{red}{ \left( \begin{matrix} 19\\ 14 \\ \end{matrix}\right )},$$
Now we write the equation of the line passing through resulting points $\color{red}{(0, 0)}$ and $\color{red}{(19, 14)}$, obtaining the correct result
$$\color{red}{14x-19y=0},$$
which is similar to your incorrect result (B).