Answer:
The solution is shown in the picture attached
Step-by-step explanation:
Answer:
[tex]D X = X^{\prime}; \ \text{therefore it is verified that} \ X \ \text{is a solution}.[/tex]
Step-by-step explanation:
[tex]\text{For} \ \ \ X = \begin{pmatrix} \sin t \\ - \frac{1}{2}\sin t - \frac{1}{2}\cos t \\ -\sin t + \cos t\end{pmatrix}[/tex]
The derivative is defined as [tex]X^{\prime} \equiv \frac{d}{dt} X = \frac{d}{dt} \begin{pmatrix} \sin t \\ - \frac{1}{2}\sin t - \frac{1}{2}\cos t \\ -\sin t + \cos t\end{pmatrix} = \begin{pmatrix} \cos t \\ -\frac{1}{2}\cos t + \frac{1}{2}\sin t \\ -\cos t - \sin t\end{pmatrix}.[/tex]
[tex]\text{If} \ \ \ \ DX = X^\prime \ \ \text{where} \ \ D = \begin{pmatrix} 1 & 0 &1 \\ 1&1&0 \\ -2&0&-1 \end{pmatrix} \ ;[/tex]
then we say [tex]X[/tex] is a solution to the given system.
Verification:
[tex]\begin{pmatrix} 1&0&1\\1&1&0\\-2&0&-1 \end{pmatrix}\begin{pmatrix} \sin t \\ -\frac{1}{2}\sin t - \frac{1}{2}\cos t \\ -\sin t + \cos t \end{pmatrix} = \begin{pmatrix} \sin t + 0 - \sin t + \cos t \\ \sin t- \frac{1}{2}\sin t - \frac{1}{2}\cos t \\ -2\sin t + \sin t - \cos t\end{pmatrix} ,[/tex]
[tex]\hspace{1cm} \Rightarrow D X = \begin{pmatrix} \cos t \\ -\frac{1}{2}\cos t + \frac{1}{2}\sin t \\ -\cos t - \sin t\end{pmatrix} = X^{\prime}.[/tex]
Hence the vector [tex]X[/tex] is a solution to the given system.
