Linear Transformation Problems
Hello, I require assistance with two LT problems.
1) Given the matrix $\begin{bmatrix} -1 &3 \\ 2 & -1 \end{bmatrix} $ , how would a circle with its center on the origin and r=1 be transformed?
2) Using the following matrix, answer the next questions. A= $\begin{bmatrix} -3 & 4 \\ 3 & 1 \end{bmatrix} $
a) Find the real numbers "a" and "b" given: A$\binom{2}{3} $ =a$\binom{2}{3} $ , A$\binom{-2}{1} $ = b$\binom{-2}{1} $
b)Find the real numbers "c, d, e and f"given: $\binom{1}{0} $ = c$\binom{2}{3} $ + d$\binom{-2}{1} $ , $\binom{0}{1} $ = e$\binom{2}{3} $ + f$\binom{-2}{1} $
c) Using the results from a), solve: $A^{n} $ $\binom{2}{3} $, $A^{n} $ $\binom{-2}{1} $
d) Using the results from b) and c), solve $A^{n} $
Answer
- The questioner was satisfied and accepted the answer, or
- The answer was disputed, but the judge evaluated it as 100% correct.
2 Attachments
Erdos
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I added some more details at the end of the file ( please see the second attachment). This was a challenging problem. Took me a while to write the solutions.
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Hey, I know I already accepted the answer, but I have a quick question if you don't mind. In the circle problem, when substituting 'y' on the equation 1, you write x=3((2X+Y)/5)-Y. Shouldn't that '-Y' be a "-X'? If that is a mistake, I can correct it myself no problem, your methodology is otherwise very clear. If it is not, could you tell me how you got that '-Y'? Thanks for all your help!
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Yes, you are right. Sorry for the mistake. It wouldn't change the result much. Just a slightly different formula for the rotated ellipse. I can fix it for you if you like. Please let me know.
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Hey, thanks for the response. Its no problem, I already did the correction, but thanks again for your support.
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Great! Sorry for my mistake again.
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