I have this matrix: $$ \overline X=\begin{pmatrix} c_1e^{-2t} +2c_2e^{5t} \\ -3c_1e^{-2t}+c_2e^{5t} \\ \end{pmatrix} $$ I want to verify that the above is a soltuion to
$$ \overline X'=\begin{pmatrix} 4 & 2 \\ 3 & -1 \\ \end{pmatrix} \overline X $$
by direct substitution:
However, I don't understand this format and can't figure out what to do. My first instinct is to take the derivative of the first matrix and substitute it into the second one, but I do not think that will work.
Any help would be helpful. Thank you!
Well, I'll do the substitution for you: $$ \begin{pmatrix} -2c_1e^{-2t} +10c_2e^{5t} \\ 6c_1e^{-2t}+ 5c_2e^{5t} \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 3 & -1 \end{pmatrix} \begin{pmatrix} c_1e^{-2t} +2c_2e^{5t} \\ -3c_1e^{-2t}+c_2e^{5t} \\ \end{pmatrix}. $$ Now you should see the format and be able to check the answer.