For the system
$x' = \left[ \begin{array}{cccc} 2&6\\3&-1 \end{array} \right]x$
with solutions
$$x_1 = \left[ \begin{array}{cccc} 2e^{5t}\\e^{5t} \end{array} \right]\qquad \text{and}\quad x_2 = \left[ \begin{array}{cccc} e^{-4t}\\-e^{-4t} \end{array} \right]$$
A. Use the Wronskian to show that the solutions are linearly independent.
B. Write the general solution to the problem.
For A, would I just compute the determinant of the matrix $x = \left[ \begin{array}{cccc} 2e^{5t}&e^{-4t}\\e^{5t}&-e^{-4t} \end{array} \right]$ and show that it never equals zero, thus proving that the two solutions are linearly independent?
For B, would the general solution just be $$x(t) = c_1\left[ \begin{array}{cccc} 2e^{5t}\\e^{5t} \end{array} \right] + c_2\left[ \begin{array}{cccc} e^{-4t}\\-e^{-4t} \end{array} \right]~?$$
Thanks!