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Linear Transformation

Question

My Attempt

I am not sure how to do 4b.. that's why I didn't try 4c as it is just the inverse of 4b.

14c14c 2015-9-22 08:11 PM

Q2你留意定義到從T(1,1)=..., T(2,3)=..., 你可以計到T(1,0)同T(0,1), 咁所以你寫得低條general formula, 咁就得, 唔洗爆matrix

Q3你個reflection寫錯左

cos(2theta) sin(2theta)
sin(2theta)  -cos(2theta)
[url]https://en.wikipedia.org/wiki/Coordinate_rotations_and_reflections[/url]

4b個做法係

T(1)=a11(1+x) + a21(x+x^2)+a31(1+x^2),

T(x^2)又係咁做, 放第三個column, 果個就係佢要搵既matrix

4c就係掉返轉, 係計T(1+x)放做a11(1)+a21(x)+a31(x^2)

[url]https://en.wikipedia.org/wiki/Linear_map[/url]

[quote]原帖由 [i]14c14c[/i] 於 2015-9-22 08:11 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=426316319&ptid=25119959][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
Q2你留意定義到從T(1,1)=..., T(2,3)=..., 你可以計到T(1,0)同T(0,1), 咁所以你寫得低條general formula, 咁就得, 唔洗爆matrix

Q3你個reflection寫錯左

Q3 You mean my rotation is wrong?

Q4b I got the matrix
1/2     3/2     -3/2
-1/2    1/2     3/2
1/2     -1/2    5/2

Q4c

1/2     -3/4     3/4
1/2     1/2        0
0        1/4      1/4

[[i] 本帖最後由 好男仔 於 2015-9-23 04:53 PM 編輯 [/i]]

edok 2015-9-23 10:53 PM

14c14c 2015-9-24 12:55 AM

[quote]原帖由 [i]好男仔[/i] 於 2015-9-23 04:13 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=426375157&ptid=25119959][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

Q3 You mean my rotation is wrong?

Q4b I got the matrix
1/2     3/2    ... [/quote]
From the question, T is a reflection across the line y=tan(theta)x for some theta
Then, the matrix should have the form
cos(2theta) sin(2theta)
sin(2theta) -cos(2theta)

one easy way to verify that your answer is wrong is that a reflection through line passing through origin should preserve length

[quote]原帖由 [i]14c14c[/i] 於 2015-9-24 12:55 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=426412274&ptid=25119959][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

From the question, T is a reflection across the line y=tan(theta)x for some theta
Then, the matrix should have the form
cos(2theta) sin(2theta)
sin(2theta) -cos(2theta)
... [/quote]

Okay.... Have to change a lot:smile_27:

So 3a, the four corners are (0,0) (3/5,4/5) (-1/5,7/5) and (-4/5,3/5)
3b (0,0) (0,1) (-1,1) and (-1,0)
3c (0,0) (-4/5,3/5) (-7/5,-1/5) and (-3/5,-4/5)
3d (0,0) (4/5,-3/5) (7/5,1/5) and (3/5,4/5)

3e
T= [-0.8 0.6; 0.6 0.8]
S= [0 -1; 1 0]
SoT=[-0.6 -0.8; -0.8 0.6]
ToS=[0.6 0.8; 0.8 -0.6]

Don't tell me its not correct:smile_27:

[quote]原帖由 [i]edok[/i] 於 2015-9-23 10:53 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=426403159&ptid=25119959][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

Thank you haha. I 比心機 becuz they contribute quite a lot to the final mark haha.
What program can I use other than Matlab? As its expensive.