# Considering the linear system Y'=AY

What would be an equation that I can use when I compute the eigenpairs for the coefficient matrix A.

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$AX=\lambda X$? –  julien May 3 at 3:52
Would I need to find the det of A –  user1988 May 3 at 3:55
oh is that where I would subtract the lambda from a and d? –  user1988 May 3 at 3:59
If you're talking about $2\times 2$ matrices with $a,d$ on the diagonal, yes. –  julien May 3 at 4:01
Yes, I should have been more specific. Thank you! –  user1988 May 3 at 4:02

Here is a start. To solve the system $Y'(t)=AY(t)$, assuming for simplicity $A_{2\times 2}$ matrix, we assume the solution to have the form
$$Y(t) = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\rm e^{\lambda t} \implies Y'(t) = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\lambda \rm e^{\lambda t}.$$
$$Y'(t) = \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\lambda \rm e^{\lambda t} = A \begin{bmatrix} k_1 \\ k_2 \end{bmatrix}\rm e^{\lambda t}\implies \begin{bmatrix} \lambda k_1 \\ \lambda k_2 \end{bmatrix}= \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} k_1 \\ k_2 \end{bmatrix}$$
and solve the system for $k_1$ and $k_2$. In order to get a non trivial solution for the system we force the determinant of the matrix of the coefficients to be zero that results in getting $\lambda's$ the eigenvalues.