To solve $x' = Ax$, where $A = \begin{bmatrix} 1 & -2 \\ 2 & 1 \end{bmatrix}$, we proceed as follows:

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Step 1: Find the Eigenvalues of $A$

The eigenvalues are solutions to the characteristic equation:

$\det(A - \lambda I) = 0$

For $A - \lambda I$:

$A - \lambda I = \begin{bmatrix} 1 - \lambda & -2 \\ 2 & 1 - \lambda \end{bmatrix}$

The determinant is:

$\det(A - \lambda I) = (1 - \lambda)(1 - \lambda) - (-2)(2)$ $\det(A - \lambda I) = (1 - \lambda)^2 - 4$ $\det(A - \lambda I) = \lambda^2 - 2\lambda - 3$

Solve for $\lambda$ by factoring:

$\lambda^2 - 2\lambda - 3 = (\lambda - 3)(\lambda + 1) = 0$

Thus, the eigenvalues are:

$\lambda_1 = 3, \quad \lambda_2 = -1$

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Step 2: Find the Eigenvectors

For each eigenvalue, find the corresponding eigenvector $v$ such that $(A - \lambda I)v = 0$.

For $\lambda_1 = 3$:

$A - 3I = \begin{bmatrix} 1 - 3 & -2 \\ 2 & 1 - 3 \end{bmatrix} = \begin{bmatrix} -2 & -2 \\ 2 & -2 \end{bmatrix}$

Solve $(A - 3I)v = 0$:

$\begin{bmatrix} -2 & -2 \\ 2 & -2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

This gives $-2v_1 - 2v_2 = 0$, or $v_1 + v_2 = 0$. Choose $v_1 = 1$, then $v_2 = -1$.

Thus, an eigenvector for $\lambda_1 = 3$ is:

$v_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$

For $\lambda_2 = -1$:

$A + I = \begin{bmatrix} 1 + 1 & -2 \\ 2 & 1 + 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix}$

Solve $(A + I)v = 0$:

$\begin{bmatrix} 2 & -2 \\ 2 & 2 \end{bmatrix} \begin{bmatrix} v_1 \\ v_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

This gives $2v_1 - 2v_2 = 0$, or $v_1 = v_2$. Choose $v_1 = 1$, then $v_2 = 1$.

Thus, an eigenvector for $\lambda_2 = -1$ is:

$v_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

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Step 3: Write the General Solution

The general solution is:

$x(t) = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2$

Substitute the eigenvalues and eigenvectors:

$x(t) = c_1 e^{3t} \begin{bmatrix} 1 \\ -1 \end{bmatrix} + c_2 e^{-t} \begin{bmatrix} 1 \\ 1 \end{bmatrix}$

Simplify:

$x(t) = c_1 \begin{bmatrix} e^{3t} \\ -e^{3t} \end{bmatrix} + c_2 \begin{bmatrix} e^{-t} \\ e^{-t} \end{bmatrix}$

Or equivalently:

$x(t) = \begin{bmatrix} c_1 e^{3t} + c_2 e^{-t} \\ -c_1 e^{3t} + c_2 e^{-t} \end{bmatrix}$

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This is the general solution to $x' = Ax$.

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Questions to Expand Understanding:

1. How does the presence of distinct eigenvalues simplify the solution process?
2. What happens if the eigenvalues are complex instead of real?
3. How would repeated eigenvalues change the solution structure?
4. Can we write this solution using matrix exponentials?
5. How do initial conditions affect the values of $c_1$ and $c_2$?

Tip:

Always verify eigenvector calculations by plugging them back into $(A - \lambda I)v = 0$ to ensure correctness.