Given \begin{equation} x' = \begin{bmatrix} 4&-3\\3&-2 \end{bmatrix}x \end{equation} What is the general solution x?
When I solve the system for the eigenvalues, its the repeated root 1; there is only one eigenvalue \begin{bmatrix} 1\\1 \end{bmatrix} so the eigenvalues are defective. I am told to find the general solution with the equation x = (ß+vt)$e^{λt}$ by first solving for ß (ß is a vector, v is the eigenvector); I get ß = v(A-tλ-1)$(λ-A)^{-1}$; however, in this case det$(λ-A)$ = 0, meaning I can't take the inverse of (λ-A) - what am I doing wrong and how do I get the general solution?