Question

Question: - Show that $\frac{\mathbf{d}}{\mathbf{d}\mathbf{s}}\left(S\overrightarrow{x}\right)\mathbf{=}\mathit{S}\frac{\mathbf{d}\overrightarrow{\mathbf{x}}}{\mathbf{d}\mathbf{t}}$.

Short answer

Answer

The solution is $\frac{\mathrm{d}}{\mathrm{ds}}\left(\mathrm{S}\overrightarrow{\mathrm{x}}\right)=S\frac{\mathrm{d}\overrightarrow{\mathrm{x}}}{\mathrm{dt}}$.

Step-by-step solution

Explanation of the solution 

Consider$\overrightarrow{x}\left(t\right)$be a differentiable curve in Rⁿ and S be an nxn matrix.

Now, consider the system as follows.

$$\begin{gathered} \frac{d}{dt}\left(S\overrightarrow{x}\right)=\frac{d}{dt}\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& \cdots & a_{nn}\end{array}\right]\left[\begin{array}{c}{x}_1\\ ⋮\\ x_n\end{array}\right] \\ =\frac{{\displaystyle d}}{{\displaystyle dt}}\left[\begin{array}{ccc}{a}_{11}{x}_1& \cdots & a_{1n}{x}_n\\ ⋮& \ddots & ⋮\\ a_{n1}{x}_1& \cdots & a_{nn}{x}_n\end{array}\right] \\ =\frac{{\displaystyle d}}{{\displaystyle dt}}\left[\begin{array}{ccc}{a}_{11}\frac{d{x}_1}{dt}& \cdots & a_{1n}\frac{d{x}_n}{dt}\\ ⋮& \ddots & ⋮\\ a_{n1}\frac{d{x}_1}{dt}& \cdots & a_{nn}\frac{d{x}_n}{dt}\end{array}\right] \\ \frac{d}{dt}\left(S\overrightarrow{x}\right)=\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& \cdots & a_{nn}\end{array}\right]\left[\begin{array}{c}\frac{d{x}_1}{dt}\\ ⋮\\ \frac{d{x}_n}{dt}\end{array}\right] \end{gathered}$$

Simplify further as follows.

$$\begin{gathered} \frac{d}{dt}\left(S\overrightarrow{x}\right)=\left[\begin{array}{ccc}{a}_{11}& \cdots & a_{1n}\\ ⋮& \ddots & ⋮\\ a_{n1}& \cdots & a_{nn}\end{array}\right]\left[\begin{array}{c}\frac{d{x}_1}{dt}\\ ⋮\\ \frac{d{x}_n}{dt}\end{array}\right] \\ \frac{d}{dt}\left(S\overrightarrow{x}\right)=S\frac{d\overrightarrow{x}}{dt} \end{gathered}$$

Hence the proof of the solution.