# Matrix, and Differential Equation Solution Verification

I given the following system:$$$$\mathbf{X'}=\begin{pmatrix}-1 & \frac14\\ 1 & -1\end{pmatrix}\mathbf{X}$$$$ Every variable in the system is a matrix. I am then given that $$\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

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

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

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 !!