Find the Image $2x-3y=0$ given matrix transformation

Question

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)

Answer 1

The line $2x-3y=0$ defines a subspace. The vector $r$ that you used is orthogonal to this line.

The correct point of view is to look at the line as set of all $(x,y)\in\mathbb R^2$ such that $(x,y)=(x, \frac{2}{3}x)$ since $y=\frac{2}{3}x$. So it suffices to check where $T$ sends, say $(3,2)$ (which is on the line). Thus, computing $\begin{pmatrix} 5& 2\\ 4& 1\end{pmatrix} \begin{bmatrix} 3\\ 2\end{bmatrix}$ should give the answer.

Answer 2

It might help to think of the problem this way: If you’ve been given some equations that describe the relationship between the pairs of variables $(x,y)$ and $(x',y')$, how might you express the equation $2x-3y=0$ in terms of $x'$ and $y'$? One way is to solve those first equations for $x$ and $y$, then substitute into $2x-3y$ and simplify. In this case, you’re given that $$\begin{bmatrix}x'\\y'\end{bmatrix} = T \begin{bmatrix}x\\y\end{bmatrix},\tag1$$ so solving for $x$ and $y$ is a simple matter of computing $T^{-1}$.

You could also work directly with vectors and matrices. The equation of the line can be written as $$\begin{bmatrix}2&-3\end{bmatrix} \begin{bmatrix}x\\y\end{bmatrix} = 0\tag2$$ (verify this). Substituting from (1) for $[x,y]^T$, this becomes $$\begin{bmatrix}2&-3\end{bmatrix}T^{-1}\begin{bmatrix}x'\\y'\end{bmatrix} = 0\tag3,$$ so the coefficients of the transformed equation are the elements of the row vector $$\begin{bmatrix}2&-3\end{bmatrix} \begin{bmatrix}5&2\\4&1\end{bmatrix}^{-1}.$$ I trust that you’re able to compute the value of this expression at this stage of your studies.

Incidentally, the vector $[2,-3]^T$ is orthogonal (normal) to the line. Equation (3) shows that such normal vectors transform differently than do vectors that represent points: If the point transformation is given by $\mathbf v' = M\mathbf v$, then a normal vector $\mathbf n$ transforms as $\mathbf n' = M^{-T}\mathbf n$. (The $-T$ superscript means “inverse transpose.”)

Answer 3

STEP 1. Realize that $\{《x, y》: 2x + 3y = 0\}$ is a subspace.

STEP 2. find the basis for the subspace. It is $(1, 2/3)$.

STEP 3. Do matrix multiplication.

$$\left[ \begin{matrix} 5 & 2 \\ 4 & 1 \\ \end{matrix}\right ] \left[ \begin{matrix} 1 \\ 2/3 \\ \end{matrix}\right ] = \left[ \begin{matrix} 19/3\\ 14/3 \\ \end{matrix}\right ]$$

STEP 4. Translate back into equation form.

$$P_1 = \alpha \left[ \begin{matrix} 19/3 \\ 4/3 \\ \end{matrix}\right ]$$ and $$P_2 = \beta \left[ \begin{matrix} 19/3 \\ 4/3 \\ \end{matrix}\right ]$$. Plug these two points (interpreting the top entries as x coordinate and bottom entries as y-coordinate). Use the point-slope equation $$y = \frac{y_2-y_1}{x_2 - x_1}x$$.

Answer 4

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).