#### xinran zhao

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##### question about a homo matrix
« on: December 02, 2018, 05:46:45 PM »
Having questions, can anyone take a look.
Question: $x'=\begin{bmatrix}-4 & 7 \\1 & -4 \end{bmatrix} x\\$
express the general solution of the given system of equation
« Last Edit: December 02, 2018, 06:06:56 PM by xinran zhao »

#### Tianyu Guo

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##### Re: question about a homo matrix
« Reply #1 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
$x'= ce^{(-4+\sqrt{7})t}*\begin{pmatrix} \sqrt{7}\\1 \end{pmatrix}+ce^{(-4-\sqrt{7})t}*\begin{pmatrix} -\sqrt{7}\\1 \end{pmatrix}$
« Last Edit: December 02, 2018, 06:25:11 PM by Tianyu Guo »

#### Ning Du

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##### Re: question about a homo matrix
« Reply #2 on: December 02, 2018, 06:16:04 PM »
$x'=\begin{bmatrix}-4 & 7 \\1 & -4 \end{bmatrix}x$
$\mathrm{Eigenvalue\:}:det\begin{bmatrix}-4-r & 7 \\1 & -4-r \end{bmatrix}=0$

$(-4-r)^2-7=0, then\ r = \sqrt{7}-4,\:-4-\sqrt{7}$

$\mathrm{Eigenvectors\:for\:}λ=\sqrt{7}-4:$

$\mathrm{Solve}(A-rI):\begin{bmatrix}-4 & 7 \\1 & -4 \end{bmatrix}-(\sqrt{7}-4)\begin{bmatrix}1 & 0 \\0 & 1 \end{bmatrix}=\begin{pmatrix}-\sqrt{7}&7\\ 1&-\sqrt{7}\end{pmatrix}$

$\mathrm{Reduce\:}\begin{pmatrix}-\sqrt{7}&7\\ 1&-\sqrt{7}\end{pmatrix}:\quad \begin{pmatrix}1&-\sqrt{7}\\ 0&0\end{pmatrix}$

$\mathrm{The\:system\:associated\:with\:the\:eigenvalue\:}λ=\sqrt{7}-4$

$\left(A-\left(\sqrt{7}-4\right)I\right)\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}1&-\sqrt{7}\\ 0&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}\\$

$\mathrm{This\:reduces\:to\:the\:equation}x-\sqrt{7}y=0,x=\sqrt{7}y\\$

$plug in and let y = 1$

$\begin{pmatrix}\sqrt{7}\\ 1\end{pmatrix}$

$\mathrm{Eigenvectors\:for\:}λ=-4-\sqrt{7}:$

$\mathrm{Solve\:}\:\left(A-\lambda\:I\right):\:\begin{pmatrix}-4&7\\ 1&-4\end{pmatrix}-\left(-4-\sqrt{7}\right)\begin{pmatrix}1&0\\ 0&1\end{pmatrix}=\begin{pmatrix}\sqrt{7}&7\\ 1&\sqrt{7}\end{pmatrix}$

$\mathrm{Reduce\:}\begin{pmatrix}\sqrt{7}&7\\ 1&\sqrt{7}\end{pmatrix}:\quad \begin{pmatrix}1&\sqrt{7}\\ 0&0\end{pmatrix}\\\mathrm{The\:system\:associated\:with\:the\:eigenvalue\:}λ=-4-\sqrt{7}$

$\left(A-\left(-4-\sqrt{7}\right)I\right)\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}1&\sqrt{7}\\ 0&0\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$

$\mathrm{This\:reduces\:to\:the\:equation}:x+\sqrt{7}y=0, x=-\sqrt{7}y$

$\mathrm{Plug\:into\:}\begin{pmatrix}x\\ y\end{pmatrix},\mathrm{Let\:}y=1, \begin{pmatrix}-\sqrt{7}\\ 1\end{pmatrix}$

$\mathrm{The\:eigenvectors\:for\:}\begin{pmatrix}-4&7\\ 1&-4\end{pmatrix}=\begin{pmatrix}\sqrt{7}\\ 1\end{pmatrix},\:\begin{pmatrix}-\sqrt{7}\\ 1\end{pmatrix}$

$x={c_1e^ \sqrt{7}t-4t}\begin{pmatrix}\sqrt{7}\\ 1\end{pmatrix}+{c_2e^ \-4t-\sqrt{7}t}\begin{pmatrix}-\sqrt{7}\\ 1\end{pmatrix}$