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I given the following system:\begin{equation}\mathbf{X'}=\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\mathbf{X}\end{equation} Every variable in the system is a matrix. I am then given that $\mathbf{X}$ is a column matrix.\begin{align}\frac{d}{dt}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}&=\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}\end{align} The excerise involves in verifing that $\mathbf{X}$ is a solution to this linear system. I am wondering if my steps are correct in assessing the problem that is given at hand.

My Steps

\begin{equation}\begin{pmatrix}\frac32e^{-\frac{3t}2}\\ -3e^{-\frac{3t}{2}}\end{pmatrix}=\begin{pmatrix}e^{-\frac{3t}2}+\frac12e^{-\frac{3t}2}\\ -e^{-\frac{3t}2}-2e^{-\frac{3t}2}\end{pmatrix}=\begin{pmatrix}\frac32e^{-\frac{3t}2}\\ -3e^{-\frac{3t}{2}}\end{pmatrix}\end{equation} $\because$ the LHS equals the RHS $\therefore$ the solution proposed by $\mathbf{X}$ is a valid one.

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  • $\begingroup$ Looks good to me ...Your steps are correct. Good job. +1 $\endgroup$ Commented Apr 18, 2020 at 1:01

1 Answer 1

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Your solution looks correct to me. $$\begin{align}\frac{d}{dt}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}&=\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\begin{pmatrix}-e^{-\frac{3t}2} \\2e^{-\frac{3t}2}\end{pmatrix}\end{align}$$ You can also write it this way: $$\begin{align}\begin{pmatrix}-1 \\ 2\end{pmatrix} \frac{d}{dt}e^{-\frac{3t}2}=&\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\begin{pmatrix}- 1\\2\end{pmatrix}e^{-\frac{3t}2}\end{align}$$ $$\begin{align}-\frac 32\begin{pmatrix}-1 \\ 2\end{pmatrix}=&\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\begin{pmatrix}- 1\\2\end{pmatrix}\end{align}$$ So that you only have matrices with numbers !!

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    $\begingroup$ You can just like manipulate the derivative on that matrices? $\endgroup$ Commented Apr 18, 2020 at 20:20
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    $\begingroup$ @EnlightenedFunky here you can simplify because you have the same exponential in each element of the matrice. You factorize the exponential. Like any constant. $\endgroup$ Commented Apr 18, 2020 at 20:34

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