Question

Write the given linear system in matrix form;

$$\begin{gathered} \left\{\frac{dx}{dt}=x-y \\ \frac{dy}{dt}=x+2z \\ \frac{dz}{dt}=-x+z\right. \end{gathered}$$

Short answer

Matrix form of the given linear system is$X\text{'}=\left(\begin{array}{ccc}1& -1& 0\\ 1& 0& 2\\ -1& 0& 1\end{array}\right)X$

Step-by-step solution

Matrix form of a Linear System

If X,A(t),and F(t) denote the respective matrices

$\mathrm{X}=\left(\begin{array}{l}{\mathrm{x}}_1\left(\mathrm{t}\right)\\ {\mathrm{x}}_2\left(\mathrm{t}\right)\\ .\\ .\\ .\\ {\mathrm{x}}_{\mathrm{n}}\left(\mathrm{t}\right)\end{array}\right),\mathrm{A}\left(\mathrm{t}\right)=\left(\begin{array}{ccc}{\mathrm{a}}_{11}\left(\mathrm{t}\right)& \dots & {\mathrm{a}}_{1\mathrm{n}}\left(\mathrm{t}\right)\\ ⋮& :& ⋮\\ {\mathrm{a}}_{\mathrm{m}1}\left(\mathrm{t}\right)& \cdots & {\mathrm{a}}_{\mathrm{mn}}\left(\mathrm{t}\right)\end{array}\right),\mathrm{F}\left(\mathrm{t}\right)=\left(\begin{array}{l}{\mathrm{f}}_1\left(\mathrm{t}\right)\\ {\mathrm{f}}_2\left(\mathrm{t}\right)\\ .\\ .\\ .\\ {\mathrm{f}}_{\mathrm{n}}\left(\mathrm{t}\right)\end{array}\right)$

Then the system of linear first order differential equations (3) can be written as,

$\frac{\mathrm{d}}{\mathrm{dt}}\left(\begin{array}{l}{\mathrm{x}}_1\left(\mathrm{t}\right)\\ {\mathrm{x}}_2\left(\mathrm{t}\right)\\ .\\ .\\ .\\ {\mathrm{x}}_{\mathrm{n}}\left(\mathrm{t}\right)\end{array}\right)=\left(\begin{array}{ccc}{\mathrm{a}}_{11}\left(\mathrm{t}\right)& \dots & {\mathrm{a}}_{1\mathrm{n}}\left(\mathrm{t}\right)\\ ⋮& :& ⋮\\ {\mathrm{a}}_{\mathrm{m}1}\left(\mathrm{t}\right)& \cdots & {\mathrm{a}}_{\mathrm{mn}}\left(\mathrm{t}\right)\end{array}\right)\left(\begin{array}{l}{\mathrm{x}}_1\\ {\mathrm{x}}_2\\ .\\ .\\ .\\ {\mathrm{x}}_{\mathrm{n}}\end{array}\right)+\left(\begin{array}{l}{\mathrm{f}}_1\left(\mathrm{t}\right)\\ {\mathrm{f}}_2\left(\mathrm{t}\right)\\ .\\ .\\ .\\ {\mathrm{f}}_{\mathrm{n}}\left(\mathrm{t}\right)\end{array}\right)$

$$" width="9" height="19" role="math">

Or simply X’=AX+F

If the system is homogenous, its matrix form is then

X’=AX

Changing the linear system in matrix form:

In the following question we will change the linear form in its matrix form;

$\begin{array}{l}\left\{\begin{array}{l}\frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{x}-\mathrm{y}\\ \frac{\mathrm{dy}}{\mathrm{dt}}=\mathrm{x}+2\mathrm{z}\\ \frac{\mathrm{dz}}{\mathrm{dt}}=-\mathrm{x}+\mathrm{z}\end{array}\right.\\ \\ \mathrm{X}\text{'}=\left(\begin{array}{ccc}1& -1& 0\\ 1& 0& 2\\ -1& 0& 1\end{array}\right)\mathrm{X}\\ \mathrm{A}=\left(\begin{array}{ccc}1& -1& 0\\ 1& 0& 2\\ -1& 0& 1\end{array}\right)\\ \\ \\ \\ \mathrm{X}\text{'}=\left(\begin{array}{ccc}1& -1& 0\\ 1& 0& 2\\ -1& 0& 1\end{array}\right)\mathrm{X}\end{array}$

Hence, the final answer is$X\text{'}=\left(\begin{array}{ccc}1& -1& 0\\ 1& 0& 2\\ -1& 0& 1\end{array}\right)X$