MHB Matrices, Eigenvalues and such...

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The discussion revolves around solving a system of differential equations represented in matrix form. The matrix A is identified as $A = \begin{pmatrix} -5 & 1 \\ 4 & -2 \end{pmatrix}$, with eigenvalues $\lambda_1 = -1$ and $\lambda_2 = -6$. Eigenvectors corresponding to these eigenvalues are found to be $\begin{bmatrix} 1 \\ 4 \end{bmatrix}$ and $\begin{bmatrix} 1 \\ -1 \end{bmatrix}$. The general solution to the system is expressed as $X(t) = C_1 \begin{bmatrix} 1 \\ 4 \end{bmatrix}e^{-t} + C_2 \begin{bmatrix} 1 \\ -1 \end{bmatrix}e^{-6t}$. A specific solution satisfying the initial conditions is derived as $X(t) = -\frac{1}{5}\begin{bmatrix} 1 \\ 4 \end{bmatrix}e^{-t} + \frac{6}{5} \begin{bmatrix} 1 \\ -1 \end{bmatrix}e^{-6t}$.
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).
 
Last edited:
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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 : $X(t) = -1/5\begin{bmatrix} 1 \\ 4 \end{bmatrix}e^{-t} + 6/5 \begin{bmatrix} 1 \\ -1 \end{bmatrix}e^{-6t}$
 
Last edited:
I'd say you've nailed it!
 

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