For the following homogeneous system of constant coefficients initial value problem, I have used a method involving matrices and confirmed that I have found the correct eigenvalues, however, I have not been successful in finding the solution to the overall initial value problem.
Consider the system of differential equations:
$$ \frac{dx}{dt} = -1.6x + 1y \\ \frac{dy}{dt} = 1.25x - 3.6y \\ x(0)= 3 \\ y(0)= 7 $$ Please see the steps of my work below. $$ \vec x= \begin{bmatrix} -1.6 && 1\\ -1.25 && -3.6\\ \end{bmatrix}x \\ $$ characteristic equation: $$ \begin{vmatrix} -1.6-r && 1\\ -1.25 && -3.6-r\\ \end{vmatrix} \\ (-3.6-r)(-1.6-r)-(1.25)(1)=0 \\ r^{2}+(\frac{26}{5})r+(\frac{451}{5})=0 \\ r_1=\frac{-41}{10}, r_2=\frac{-11}{10} $$
for eigenvalue $r_1$:
$$ \left[ \begin{array}{cc|c} -1.6-(\frac{-41}{10}) & 1 & 0\\ -1.25 & -3.6-(\frac{-41}{10}) & 0\\ \end{array} \right] \\ \left[ \begin{array}{cc|c} 2.5 & 1 & 0\\ -1.25 & 0.5 & 0\\ \end{array} \right] $$ RREF $$ \left[ \begin{array}{cc|c} 1 & 0.4 & 0\\ 0 & 0 & 0\\ \end{array} \right] \\ \xi_1+0.4\xi_2=0 \\ $$ let: $$ \xi_1=1,\xi_2=\frac{-4}{10} $$ for eigenvalue $r_2$:
$$ \left[ \begin{array}{cc|c} -1.6-(\frac{-11}{10}) & 1 & 0\\ -1.25 & -3.6-(\frac{-11}{10}) & 0\\ \end{array} \right] \\ \left[ \begin{array}{cc|c} -\frac{1}{2} & 1 & 0\\ \frac{5}{4} & -\frac{5}{2} & 0\\ \end{array} \right] $$ RREF $$ \left[ \begin{array}{cc|c} 1 & -2 & 0\\ 0 & 0 & 0\\ \end{array} \right] \\ \xi_1-2\xi_2=0 \\ $$ let: $$ \xi_1=1,\xi_2=2 $$ General Solution: $$ x(t)=c_1e^{-\frac{41}{10}t} \begin{pmatrix} -\frac{4}{10}\\ 1\\ \end{pmatrix}+c_2e^{-\frac{11}{10}t}\begin{pmatrix} 2\\ 1\\ \end{pmatrix} $$ Solving for particular solution by applying initial conditions: $$ \left[ \begin{array}{cc|c} -\frac{4}{10} & 2 & 3\\ 1 & 1 & 7\\ \end{array} \right] $$ RREF: $$ \left[ \begin{array}{cc|c} 1 & 0 & \frac{55}{12}\\ 0 & 1 & \frac{29}{12}\\ \end{array} \right] \\ \begin{pmatrix} c_1\\ c_2\\ \end{pmatrix}= \begin{pmatrix} \frac{55}{12}\\ \frac{29}{12}\\ \end{pmatrix} \\ \vec{x}(t)= \begin{bmatrix} -\frac{11}{6}e^{-\frac{41}{10}t}+\frac{29}{6}e^{-\frac{11}{10}t}\\ \frac{55}{12}e^{-\frac{41}{10}t}+\frac{29}{12}e^{-\frac{11}{10}t}\\ \end{bmatrix} $$