Fdtxjr

Be $M$ the linear transformation represented by matrix $T$:

$$T =
left[ beginmatrix
5 & 2 \
4 & 1 \
endmatrixright ]
$$

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: $vecr = left[beginmatrix 2\-3endmatrixright]$

$vecz = T . vecr = left[beginmatrix 4\-5endmatrixright]$ (wrong)

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 $(x,y)inmathbb R^2$ such that $(x,y)=(x, frac23x)$ since $y=frac23x$. So it suffices to check where $T$ sends, say $(3,2)$ (which is on the line). Thus, computing
$beginpmatrix 5& 2\ 4& 1endpmatrix beginbmatrix 3\ 2endbmatrix$
should give the 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

$$left( beginmatrix
5 & 2 \
4 & 1 \
endmatrixright )cdot
left( beginmatrix
0 \
0 \
endmatrixright ) = colorred
left( beginmatrix
0\
0 \
endmatrixright ),quadleft( beginmatrix
5 & 2 \
4 & 1 \
endmatrixright )cdot
left( beginmatrix
3 \
2 \
endmatrixright ) = colorred
left( beginmatrix
19\
14 \
endmatrixright ),$$

Now we write the equation of the line passing through resulting points $colorred(0, 0)$ and $colorred(19, 14)$, obtaining the correct result

$$colorred14x-19y=0,$$

which is similar to your incorrect result (B).

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 $$beginbmatrixx'\y'endbmatrix = T beginbmatrixx\yendbmatrix,tag1$$ 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 $$beginbmatrix2&-3endbmatrix beginbmatrixx\yendbmatrix = 0tag2$$ (verify this). Substituting from (1) for $[x,y]^T$, this becomes $$beginbmatrix2&-3endbmatrixT^-1beginbmatrixx'\y'endbmatrix = 0tag3,$$ so the coefficients of the transformed equation are the elements of the row vector $$beginbmatrix2&-3endbmatrix beginbmatrix5&2\4&1endbmatrix^-1.$$ 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 $mathbf v' = Mmathbf v$, then a normal vector $mathbf n$ transforms as $mathbf n' = M^-Tmathbf n$. (The $-T$ superscript means “inverse transpose.”)

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.

$$left[ beginmatrix
5 & 2 \
4 & 1 \
endmatrixright ]
left[ beginmatrix
1 \
2/3 \
endmatrixright ] =
left[ beginmatrix
19/3\
14/3 \
endmatrixright ]$$

STEP 4. Translate back into equation form.

$$P_1 = alpha left[ beginmatrix
19/3 \
4/3 \
endmatrixright ]$$ and $$P_2 = beta left[ beginmatrix
19/3 \
4/3 \
endmatrixright ]$$. Plug these two points (interpreting the top entries as x coordinate and bottom entries as y-coordinate). Use the point-slope equation $$y = fracy_2-y_1x_2 - x_1x$$.