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MAT244--Lectures & Home Assignments / Re: question about a homo matrix
« on: December 02, 2018, 06:08:46 PM »
We can compute the eigenvalues first
Let $A=\begin{pmatrix}-4~~~~7\\1 ~~~-4 \end{pmatrix}$
Setting up $det\begin{pmatrix} -4-\lambda ~~~7 \\ 1 ~~~-4-\lambda \end{pmatrix} = 0$
then we find out $(-4-\lambda)^2 -7=0$
By solving the equation, we get the eigenvalue $r_1=-4+\sqrt{7}$ and $r_2=-4-\sqrt{7}$.
Now, we need to plug the eigenvalue back to the matrix.
For $r_1=-4+\sqrt{7}$, the corresponding eigenvector is $\begin{pmatrix} \sqrt{7}\\1 \end{pmatrix}$.
For $r_2=-4-\sqrt{7}$, the corresponding eigenvector is $\begin{pmatrix} -\sqrt{7}\\1 \end{pmatrix}$
Thus, the general solution for the system of equation is
$x'= ce^{(-4+\sqrt{7})t}*\begin{pmatrix} \sqrt{7}\\1 \end{pmatrix}+ce^{(-4-\sqrt{7})t}*\begin{pmatrix} -\sqrt{7}\\1 \end{pmatrix}$
Let $A=\begin{pmatrix}-4~~~~7\\1 ~~~-4 \end{pmatrix}$
Setting up $det\begin{pmatrix} -4-\lambda ~~~7 \\ 1 ~~~-4-\lambda \end{pmatrix} = 0$
then we find out $(-4-\lambda)^2 -7=0$
By solving the equation, we get the eigenvalue $r_1=-4+\sqrt{7}$ and $r_2=-4-\sqrt{7}$.
Now, we need to plug the eigenvalue back to the matrix.
For $r_1=-4+\sqrt{7}$, the corresponding eigenvector is $\begin{pmatrix} \sqrt{7}\\1 \end{pmatrix}$.
For $r_2=-4-\sqrt{7}$, the corresponding eigenvector is $\begin{pmatrix} -\sqrt{7}\\1 \end{pmatrix}$
Thus, the general solution for the system of equation is
$x'= ce^{(-4+\sqrt{7})t}*\begin{pmatrix} \sqrt{7}\\1 \end{pmatrix}+ce^{(-4-\sqrt{7})t}*\begin{pmatrix} -\sqrt{7}\\1 \end{pmatrix}$