Question

Find a general solution of the system.

Answer 1

Care to write the system? .. o.o

Answer 2

\(X'(t)=\begin{bmatrix}
1 &0 &-1 \\
0 &2 &0 \\
1 &0 &1
\end{bmatrix}X(t)+\begin{bmatrix}
-1\\
-1-e^{-t}\\
-2e^{-t}\\
\end{bmatrix}\)

Sorry, kinda Latex intensive to write.

Answer 3

Oh, well..matrices are something i'm not good at, sorry.

Answer 4

I can do it for you, but I need your help, exchange, deal?

Answer 5

Sure, I can sure try and help you.

Answer 6

I am kicked out of the net, sorry, now I work on it.

Answer 7

want to do with me ?

Answer 8

Well, I can... but I am working on like 3 problems at once right now actually lol.

Answer 9

ok, I work and will post after finish. you come here to help me. http://openstudy.com/study#/updates/52a100cae4b0d45cde225005 Ok

Answer 10

are you going to show work as well?

Answer 11

I'll try all my best.

Answer 12

take a look at this first, if you don't get something, ask me then. Still have partial part

Answer 13

sorry, cost , not cos x/

Answer 14

Your roots are wrong, I think you should get 2,2,1 for the r's though.

Answer 15

ok, let me check, I did on calculator

Answer 16

I think it will end up being

(r-2)(r^2-3r+2) for the determinant.

Answer 17

I don't know how can you get that equation, I can say: mine is 100% correct. I double check

Answer 18

the characteristic equation is r^3 -4r^2+6r-4
while yours is r^3 -5r^2 +8r -4
why we don't meet to each other?

Answer 19

\(\begin{bmatrix}
1 &0 &-1 \\
0 &2 &0 \\
1 &0 &1
\end{bmatrix}-\begin{bmatrix}
r &0 &0 \\
0 &r &0 \\
0 &0 &r
\end{bmatrix}=\begin{bmatrix}
1-r &0 &-1 \\
0 &2-r &0 \\
1 &0 &1-r
\end{bmatrix}\)

From that,

\(determinant\rightarrow (1-r)\begin{bmatrix}
2-r&0\\
0&1-r
\end{bmatrix}-0+(-1)\begin{bmatrix}
0&2-r\\
1&0
\end{bmatrix}\)

Which then simplifies down to,

\((r-2)(r^2-3r+2)\)

Which factors to,
\((r-2)(r-2)(r-1)\)

This is what I have.

Answer 20

so, mine is ok,

Answer 21

should be -3r in the middle :)

Answer 22

hello, we have to meet to each other before we step up. right? My way in finding characteristic equation is not use determinant but it is always correct.

Answer 23

(1-r)^2 = 1^2 -2r +r^2 , where do you have 3r?

Answer 24

have to go to eat something, you work , I'll be back

Answer 25

I will just go with your way...

Answer 26

hello, are you there? meet me or else? any conflict?

Answer 27

I am fairly confident you have the wrong r values, but I should at least get partial credit.

Answer 28

no, we Must meet to each other, if not, the leftover cannot be correct since it base on this part.
I never use determinant method to construct characteristic equation for 3 x 3 matrix, because it's easy to make mistake while my way is rarely

Answer 29

could you rework it with r values of r=2,2,1 ?

Answer 30

@austinL I confirm. http://openstudy.com/study#/updates/52a1274ce4b0a13bbbe89c51
if you agree with me, we can step up. if not, I give up