Transformations matrices have always been sort of a weak area for me just because it always feels like I'm doing the process wrong.

The question I'm trying to solve is this:

Suppose we stretch the picture horizontally by a factor of 2, rotate it 45 degrees clockwise, shrink it horizontally by a factor of 3 and then reflect it through the y-axis

My attempt at the question:

The first step's matrix would be: \begin{bmatrix}2&0\\0&1\end{bmatrix}

The second step's matrix would be: \begin{bmatrix}\cos(-\pi/4)&-\sin(-\pi/4)\\\sin(-\pi/4)&\cos(-\pi/4)\end{bmatrix}

The third step's matrix would be: \begin{bmatrix}1/3&0\\0&1\end{bmatrix}

The last step's matrix would be: \begin{bmatrix}-1&0\\0&1\end{bmatrix}

I multiplied all the transformation matrices together in the order given and got a final answer of approximately: \begin{bmatrix}-0.4714&1.4142\\0.2357&0.7071\end{bmatrix}

I just want to make sure what I did was correct. If I made an error somewhere in the middle, please let me know. I cannot understand transformation matrices for the life of me :(

  • 1
    $\begingroup$ If the transforms as listed are A, B, C, D then you want DCBA. Your matrices look fine but I don't get the same final product as you $\endgroup$
    – Paul
    Feb 21 '20 at 9:56
  • $\begingroup$ May I ask what you mean by "you want DCBA"? $\endgroup$ Feb 21 '20 at 16:57
  • 1
    $\begingroup$ You multiply in that order, not ABCD. The first transformation is on the right, not the left. $\endgroup$
    – Paul
    Feb 21 '20 at 19:15

Have you heard of 3blue1brown? That YouTube channel is the only reason I understood linear algebra and more.

Basically all you do is follow where the basis vectors go. Very smoothly and patiently, you will get an answer.

https://www.youtube.com/playlist?list=PLZHQObOWTQDPD3MizzM2xVFitgF8hE_ab https://youtu.be/XkY2DOUCWMU

But if you want to confirm your math homework answers very quickly you could try this Desmos feature I dug up.


I won't compute the actual thing here cause Latex is stressful on a phone. But if this down here is what you worked out, then you'll be fine(matrix multiplication is inside out like function composition):

$\begin{bmatrix} -1 & 0\\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{1}{3} & 0\\ 0 & 1 \end{bmatrix} \begin{bmatrix} \frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}\\ -\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2} \end{bmatrix} \begin{bmatrix} 2 & 0\\ 0 & 1 \end{bmatrix}$

Don't despair so easily. With enough practice, patience and intuition in something especially sweet like linear algebra you will get better at math.

Edit: I'm not sure how you did the multiplication cause I got:

$\begin{bmatrix} -\frac{\sqrt{2}}{3} & -\frac{\sqrt{2}}{6}\\ -\sqrt{2} & \frac{\sqrt{2}}{2} \end{bmatrix}$

  • $\begingroup$ How did you get one of your matrices as sqrt2/2? Why is it not 1/sqrt2? $\endgroup$ Feb 21 '20 at 16:57
  • $\begingroup$ Oh it's the same thing. Influence of my Physics teacher. That's how I always write it. But to be clear $\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}$ $\endgroup$ Feb 21 '20 at 17:00
  • $\begingroup$ Ah I see. That makes sense. So is it that my matrices were correct but my multiplication was done wrong? Is it right that the multiplication is step1 * step2 * step3* step4? $\endgroup$ Feb 21 '20 at 17:06
  • $\begingroup$ Yes. If the picture was operated on like that, you have to take it as it is. But I really recommend you check out those links. By just following $\hat \imath, \hat \jmath$ I was able to do it in 3 minutes only. $\endgroup$ Feb 21 '20 at 17:13
  • $\begingroup$ Sounds good thank you for your help $\endgroup$ Feb 21 '20 at 17:14

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