Question

Write the given linear system in matrix form.

$$\begin{gathered} \frac{dx}{dt}=x-y+z+t-1 \\ \frac{dy}{dt}=2x+y-z-3t^2 \\ \frac{dy}{dt}=x+y+z+t^2-t+2 \end{gathered}$$

Short answer

The matrix is

$X\text{'}=\left(\begin{array}{ccc}1& -1& 1\\ 2& 1& -1\\ 1& 1& 1\end{array}\right)X+\left(\begin{array}{c}0\\ -3\\ 1\end{array}\right)+t\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)+t+\left(\begin{array}{c}-1\\ 0\\ 2\end{array}\right)$

Step-by-step solution

Analysing the question:

Writing in matrix form;

For$\mathrm{X}=\left(\begin{array}{c}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right)$matrix notation of system is

$\mathrm{X}\text{'}=\left(\begin{array}{ccc}1& -1& 1\\ 2& 1& -1\\ 1& 1& 1\end{array}\right)\mathrm{X}+\left(\begin{array}{c}0\\ -3\\ 1\end{array}\right)\mathrm{t}+\left(\begin{array}{c}-1\\ 0\\ 2\end{array}\right)$

Where

$\mathrm{A}=\left(\begin{array}{ccc}1& -1& 1\\ 2& 1& -1\\ 1& 1& 1\end{array}\right)$

Hence, the final answer is $\mathrm{X}\text{'}=\left(\begin{array}{ccc}1& -1& 1\\ 2& 1& -1\\ 1& 1& 1\end{array}\right)\mathrm{X}+\left(\begin{array}{c}0\\ -3\\ 1\end{array}\right)+\mathrm{t}\left(\begin{array}{c}1\\ 0\\ -1\end{array}\right)+\mathrm{t}+\left(\begin{array}{c}-1\\ 0\\ 2\end{array}\right)$