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Mathematics 22 Online
OpenStudy (decentnabeel):

Please Help ! Medal and fan!

OpenStudy (decentnabeel):

\[\left[\begin{matrix}3 & 1 \\ 2 & 4\end{matrix}\right]\] \[z=\frac{3}{5}\] Calculate the A^n.Z

OpenStudy (puertoricangirl):

I will try

OpenStudy (fwizbang):

What's n? Z?

OpenStudy (decentnabeel):

Z=3/5

OpenStudy (fwizbang):

n=?

OpenStudy (decentnabeel):

OpenStudy (fwizbang):

Z is a column vector. \[Z=\left(\begin{matrix}3 \\ 5\end{matrix}\right)\] The line is a stray mark on he paper not a fraction. Try multiplying Z by A once and compare the answer to Z. What happens?

OpenStudy (decentnabeel):

A ^n multiple by Z

OpenStudy (decentnabeel):

so how to find n

OpenStudy (fwizbang):

I know, humor me....

OpenStudy (fwizbang):

If YOU multiply AZ, what do you get?

OpenStudy (fwizbang):

Oops, make that if you multiply AZ, what do you get?

OpenStudy (decentnabeel):

@fwizbang Az=\[\left(\begin{matrix}14 \\ 26\end{matrix}\right)\]

OpenStudy (decentnabeel):

@mathmale

OpenStudy (fwizbang):

Ok, now I see what's happening. We need to write Z as a linear combination of X and Y(which are eigenvectors of A.) So, find an a and a b such that Z=aX+bY

OpenStudy (decentnabeel):

yes

OpenStudy (fwizbang):

Then if we now the eigenvalues, we can write the answer for a general integer n. If\[AX=\lambda _{x}X\] then \[A^nX=(\lambda _{x})^n X\]

OpenStudy (fwizbang):

Same idea for Y, so do yu know the eigenvalues for X and Y?

OpenStudy (decentnabeel):

no

OpenStudy (fwizbang):

Ok. Multiply X by A and compare the answer to X. Same for Y.

OpenStudy (fwizbang):

It looks like you already did this on your sheet....

OpenStudy (decentnabeel):

Multiply X by A \[\left(\begin{matrix}2 \\ -2\end{matrix}\right)\]

OpenStudy (fwizbang):

How does that compare to X?

OpenStudy (decentnabeel):

\[\left(\begin{matrix}2 \\ -2\end{matrix}\right)=\left(\begin{matrix}1 \\ -1\end{matrix}\right)\]

OpenStudy (fwizbang):

x 2 on the right? So the eigenvalue for X is 2.

OpenStudy (fwizbang):

Now do Y....

OpenStudy (decentnabeel):

Thanks can you help one more Question

OpenStudy (fwizbang):

Sure. What do you got?

OpenStudy (decentnabeel):

-2y^2=4x Find the directrix and the focus of the parabola

OpenStudy (decentnabeel):

@fwizbang

OpenStudy (fwizbang):

So if you can write the equation as \[(y-k)^2=4p(x-h)\] then the focus it at (h+p,k) and the directrix is x=h-p.( I just googled parabola)

OpenStudy (decentnabeel):

-2y^2=4x divided -2 on both sides y^2=-2x so then

OpenStudy (fwizbang):

So 4p=-2, p=? PS (y-0)^2=y^2.

OpenStudy (decentnabeel):

p=-1/2 focus is(-1/2,0) Right

OpenStudy (fwizbang):

yes!

OpenStudy (decentnabeel):

Directis=-1/2

OpenStudy (fwizbang):

Careful h minus p.....

OpenStudy (decentnabeel):

0-(-1/2)=1/2 so that directix =1/2

OpenStudy (fwizbang):

Now you got it.

OpenStudy (decentnabeel):

yes thanks :) @fwizbang

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