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.
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 !!