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+2
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avatar+700 

The problem states that this matrix: 

\(\begin{bmatrix} 0 && -1 \\ 1&& 0 \end{bmatrix}\)

 

 

is multiplied by this matrix: (these are the vertices of a parallelogram btw the first row represents the x coordinate the second represents the y coordinate and the colums each represent a point: a,b,c...etc.
\(\begin{bmatrix} 5 &3&0& -2 \\ -2 &2&0&4 \end{bmatrix}\)


However when I did the work.....
This was my resulting matrix:
\(\begin{bmatrix} 2&-2&0&-4\\ 5&3&0&-2 \end{bmatrix}\)

 

 

The key however when i came to check my answer said the answer was some other matrix and that the parallelogram had this transformation: 
a 270 degree rotation about the origin

However I said: 
a 90 degree rotation about the origin.....


Did i do something wrong? if yes, can someone please explain what i did wrong :0 

thank you so much! 

 Nov 15, 2019
 #1
avatar+2863 
0

Oof haven't learned matrices yet...

 

Welp I can only solve some competition problems,

 Nov 15, 2019
 #2
avatar+129852 
+2

Your resulting matrix is correct

 

A 90 °  rotation clockwise  about the origin  has  the tranformation formula :

 

(x, y) ⇒ (  y, -x)

So

(5 , -2) ⇒ ( -2, -5)     which isn't correct

 

A  270° rotation clockwise about the origin  has  the transformation formula  :

 

   ( x , y) ⇒  (-y , x)

So

(5, -2) ⇒  ( 2 , 5)

(3, 2) ⇒  (- 2, 3)

(0, 0 ) ⇒ (0, 0)

(-2, 4)  ⇒ ( -4 , -2)

 

Here are the transformation rules, Nirvana :

 

https://quizlet.com/74135624/rotation-counterclockwise-reflection-coordinate-rules-flash-cards/

 

 

cool cool cool

 Nov 15, 2019
 #3
avatar+700 
+1

OK. Thank you, but I have a question.......since they told us to multiply by \(\begin{bmatrix} 0&& -1 \\ 1&& 0 \end{bmatrix}\)

shouldn't it be 90 degrees? 

Nirvana  Nov 15, 2019
 #4
avatar+129852 
+2

The matrix for  a 90°  counter-clockwise rotation  is

 

[ 0   -1  ]

[ 1    0  ]

 

But this is the same as a 270° clockwise rotation

 

See the rotation matrices here : https://en.wikipedia.org/wiki/Rotation_matrix

 

cool cool cool

 Nov 15, 2019
edited by CPhill  Nov 15, 2019
 #5
avatar+700 
+1

oh! 
wait so now im confused........am i right or wrong.....

Nirvana  Nov 15, 2019
 #6
avatar+129852 
+1

Your answer was incorrect

 

A 90 °  rotation clockwise  about the origin  has  the tranformation formula :

 

(x, y) ⇒ (  y, -x)

 

So

 

(5, -2)  would become ( -2, -5)

(3,2)  would become (2, -3)

(0,0)  would become (0, 0)

(-2, -4) would become ( -4, 2)

 

But your  transformed  points should be for a -90°  rotation  [ =  a 270°  rotation ]

 

(2,5)   (-2,3)  (0, 0)   and  ( -4, -2)

 

Which  are the points on your transformed matrix

 

 

cool cool cool

CPhill  Nov 15, 2019
edited by CPhill  Nov 15, 2019
 #7
avatar+700 
0

on my paper i said 90 degrees counterclockwise which.............i thought that 90 degrees clockwise meant 270 degrees ccw and -90 meant 270 ccw 

woah....what im so confused im so sorry )::::

Nirvana  Nov 15, 2019
edited by Nirvana  Nov 15, 2019
edited by Nirvana  Nov 15, 2019
 #8
avatar+129852 
+2

See if this helps :

 

 

So    a 90°  counter-clockwise rotation  would put the vector  pointing up the y axis in a positive direction

 

The rotational matrix for  a 90° counter-clockwise rotation is the one you were given 

 

But....this is the same thing as a 270°  clockwise rotation

 

  Does this make sense  ???

 

cool cool cool

 Nov 15, 2019
edited by CPhill  Nov 15, 2019
 #9
avatar+700 
+1

yes yes!! thank you so much CPhill!

Nirvana  Nov 15, 2019

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