# 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 $A^{n}$

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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.

• 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!

• 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.

• Hey, thanks for the response. Its no problem, I already did the correction, but thanks again for your support.

• Great! Sorry for my mistake again.