Question

Write the given linear system in matrix form.

$$\begin{gathered} \frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{4}\mathit{x}\mathbf{-}\mathbf{7}\mathit{y} \\ \frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{t}}\mathbf{=}\mathbf{5}\mathit{x} \end{gathered}$$

Short answer

The matrix is$X\text{'}=\left(\begin{array}{cc}4& -5\\ 5& 0\end{array}\right)X$

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,

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

Writing in matrix form;

The matrix notation is given by X’=AX. Therefore the above equation can be written as,

$\begin{array}{rcl}\left(\begin{array}{c}{x}^{,}\\ y^{,}\end{array}\right)& =& \left(\begin{array}{cc}4& -5\\ 5& 0\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)\\ A& =& \left(\begin{array}{cc}4& -5\\ 5& 0\end{array}\right)\\ X^{,}& =& \left(\begin{array}{cc}4& -5\\ 5& 0\end{array}\right)X\end{array}$

Hence, the final answer is $X^{,}=\left(\begin{array}{cc}4& -5\\ 5& 0\end{array}\right)X$