MHB Matrices, Eigenvalues and such...

shamieh
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Consider the system

$x'_1 = -5x_1 + 1x_2$
$x'_2 = 4x_1 - 2x_2$

If we write in matrix form as $X' = AX$ then

a) X =

b) X' =

c) A =

d) Find the eigenvalues of A.

e) Find eigenvectors associated with each eigenvalue. Indicate which eigenvector goes with which eigenvalue.

f) Write the general solution to the system.

g) Find the specific solution that satisfies the initial conditions $x_1(0) = 1$ and $x_2(0) = -2$

So I am not really sure on how to solve these...Here is what I have so far
My Solutions

a) X= $\overrightarrow{X} = (^{x_1} _{x_2})$

b) X' = $(^{-5x_1 + 1x_2}_{4x_1 - 2x_2})$

c) A = $(^{-5}_4$ $^1_{-2})$

d) eig values of A : $\lambda_1 = -1$ and $\lambda_2 = -6$

e) help
f) help
g) refer to e and f (help).
 
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I agree with your eigenvalues. To find the eigenvectors, you need to solve the system $(A-\lambda I)x_{\lambda}=0$ for $x_{\lambda}$, which is your eigenvector. You should do this separately, once for each eigenvalue. Each system should be degenerate. What do you get?
 
e) so for the eigenvectors i got: \begin{bmatrix} 1 \\ 4 \end{bmatrix} and \begin{bmatrix} 1 \\ -1 \end{bmatrix}

for part f) $X(t) = C_1 \begin{bmatrix} 1 \\ 4 \end{bmatrix}e^{-t} + C_2 \begin{bmatrix} 1 \\ -1 \end{bmatrix}e^{-6t}$

and for part g) I got : $X(t) = -1/5\begin{bmatrix} 1 \\ 4 \end{bmatrix}e^{-t} + 6/5 \begin{bmatrix} 1 \\ -1 \end{bmatrix}e^{-6t}$
 
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I'd say you've nailed it!
 
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