Question

# Find a fundamental matrix for the system and write the general solution as a matrix. If initial values are given, solve the initial value problem. $x_{1}^{\prime}=2 x_{1}+x_{2}-x_{3}, x_{2}^{\prime}=3 x_{1}-2 x_{2}, \ x_{3}^{\prime}=3 x_{1}+x_{2}-3 x_{3} ; x_{1}(0)=1, x_{2}(0)=7, x_{3}(0)=3$

Solution

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Let

\begin{align*}\mathbf{X}=\begin{pmatrix} x_1\\ x_2\\x_3 \end{pmatrix}\quad \text{ and }\quad \mathbf{X}'=\begin{pmatrix} x_1'\\ x_2'\\x_3' \end{pmatrix} \end{align*}

Let's rewrite the given system in terms of matrices:

$\mathbf{X}'=\mathbf{AX}$

with $\mathbf{A}$ being a real, constant, $3\times 3$ matrix.

\begin{align*}\mathbf{X}'=\begin{pmatrix} x_1'\\ x_2'\\x_3'\end{pmatrix}&=\mqty(2x_1+x_2-x_3 \\ 3x_1-2x_2\\3x_1+x_2-3x_3)=\mqty (2&\phantom{ii}1&-1\\3&-2&\phantom{ii}0\\ 3&\phantom{ii}1&-3)\begin{pmatrix} x_1\\ x_2\\x_3 \end{pmatrix} \end{align*}

Thus,

\begin{align*}\mathbf{X}'=\underbrace{\mqty (2&\phantom{ii}1&-1\\3&-2&\phantom{ii}0\\ 3&\phantom{ii}1&-3)}_{\mathbf{A}} \mathbf{X} \end{align*}

For our example,

\begin{align*}\mathbf{A}=\mqty (2&\phantom{ii}1&-1\\3&-2&\phantom{ii}0\\ 3&\phantom{ii}1&-3)\end{align*}

The solution $\mathbf{X}$ is given in the form $\mathbf{X=E}e^{\lambda t}$ where $\lambda$ is an eigenvalue of $\mathbf{A}$ with associated eigenvector $\mathbf{E}$.

$\mathbf{AE=}\lambda \mathbf{E}$

Recall, $\lambda$ is an $\textbf{eigenvalue}$ of $\mathbf{A}$ if and only if it's a root of charateristic polynomial of $\mathbf{A}$. $\lambda$ is obtain by calculating

$p_A(\lambda)=|\lambda \mathbf{I_n}-\mathbf{A}|=0$

The degree of $p_A(\lambda )$ determines the number of eigenvalues.Any nontrivial solution $\mathbf{E}$ of

$(\lambda \mathbf{I}_n-\mathbf{A})\mathbf{X}=0$

is an eigenvalue of $\mathbf{A}$ associated with $\lambda$.

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