I have the following problem:
$$A=\begin{bmatrix}-9 & 3\\ 3 & -1\end{bmatrix}, f(t)=\begin{bmatrix}t^{-1}\\7+3t^{-1}\end{bmatrix}$$
Solving for the eigenvalues of A, I get -10, 0.
Using $r=10$ and completing row reduction, I get: $$\begin{bmatrix}1 & -3\\ 0 & 0\end{bmatrix}, \begin{bmatrix}u_1\\u_2\end{bmatrix}=\begin{bmatrix}-3\\1 \end{bmatrix}$$
Using $r=0$ and completing row reduction, I get: $$\begin{bmatrix}1 & -\frac{1}{3}\\ 0 & 0\end{bmatrix}, \begin{bmatrix}u_1\\u_2\end{bmatrix}=\begin{bmatrix}1\\3 \end{bmatrix}$$
From this, I get that $$X=\begin{bmatrix}-3e^{-10t} & 1\\ e^{-10t} & 3\end{bmatrix}, X^{-1}=\begin{bmatrix}-\frac{3}{10}e^{10t} & \frac{1}{10}e^{10t}\\ \frac{1}{10} & \frac{3}{10}\end{bmatrix}$$
$$X^{-1}f=\begin{bmatrix}\frac{\frac{3}{t}+7}{10}e^{10t}-\frac{3e^{10t}}{10t} \\ \frac{1}{t}+\frac{21}{10}\end{bmatrix}$$
Taking the integral of that, I get: $$\begin{bmatrix}\frac{7e^{10t}}{100} \\ \ln{|t|}+\frac{21t}{10}\end{bmatrix}$$
Multiplying $X*\int X^{-1}f(t) \, dx$, I get: $$\begin{bmatrix}\frac{21}{10}t+\ln{|t|}+\frac{21}{100} \\ \frac{63}{10}t+3\ln{|t|}+\frac{7}{100}\end{bmatrix}$$
Putting together the whole answer, I then arrive at: $$x(t)=c_1e^{-10t}\begin{bmatrix}-3\\1\end{bmatrix}+c_2\begin{bmatrix}1\\3\end{bmatrix}+\begin{bmatrix}\frac{21}{10}t+\ln{|t|}+\frac{21}{100} \\ \frac{63}{10}t+3\ln{|t|}+\frac{7}{100}\end{bmatrix}$$
Which is not correct. I'm not sure on what step I'm going wrong, I was able to do the other handful of problems correct with this method.