1) Find eigenvalues for the matrix A=((3,-2),(2,-2)). Here and later first inner parentheses denote first row, second--second row. In order to do it, one need to solve secular equation det|A-k*I|=0, where k is the eigenvalue to be found and I is unit matrix, I=((1,0),(0,1)). In our case we will obtain:

det|((3-k,-2),(2,-2-k))|=0. From this we will get:

(3-k)(-2-k)-2*(-2)=0 or (k-3)*(k+2)+4=0 or k^{2}-k-2=0

Solutions of this quadratic equation are given by:

k_{1}=1/2*[1+√(1-4*(-2))]=(1+√9)/2=2

k_{2}=1/2*[1-√(1-4*(-2))]=(1-√9)/2=-1

2) Now, let us find eigenvectors corresponding to each of the eigenvalues. We plug in first eigenvalue, k=2, into the matrix equation above to obtain:

K_{1}=A-k_{1}*I=A-2I=((1,-2),(2,-4)); Now we need to solve K_{1}(a_{1},a_{2})=0 for vector (a_{1},a_{2}). If we use the explicit form of matrix K_{1}, we will obtain two identical equations:

a_{1}-2a_{2}=0 and 2a_{1}-4a_{2}=0. Those have infinitely many solutions, corresponding to infinitely many vectors, which all point in the same direction, but differ in length. Let us pick the most obvious solution, a_{1}=2; a_{2}=1; Thus, first eigenvector is (2,1). Often eigenvectors are normalized, then this eigenvector would be (2,1)/√5.

Next, we do the same for the other eigenvalue, k_{2}. Upon substitution and solving the equation, we will obtain the second eigenvector, (1,2).

3) General solution to a differential equation in question is in the form:

C_{1}*(first eigenvector)*Exp[k_{1}t]+C_{2}*(second eigenvector)*Exp[k_{2}t]

Here k_{1} shall correspond to its eigenvector and so shall k_{2}! Do not mix them up!

In our case, (2,1) eigenvector corresponds to k_{1}=2 and (1,2) eigenvector corresponds to k_{2}=-1, thus we obtain:

x(t)=C_{1}*(2,1)*e^{2t}+C_{2}*(1,2)*e^{-t};