Example 4: Solve the system .

Solution: As before, we define the matrix , the vector and the identity matrix .

> A:=matrix(2,2,[1,- 1,1, 3]);

> t:=matrix(2,1,[t1,t2]);

> E:=array(1..2,1..2,identity);

Next we find eigenvalues of .

> eigenvalues(A);

Notice that the eigenvalue 2 is a repeated eigenvalue, which will yield only one independent solution. We find this eigenvector in the same way we did before.

> multiply(A-2*E,t);

> solve(-t1-t2=0,t2);

Hence one eigenvector is which gives one independent solution . We find a second solution as follows:

First solve the matrix equation .

> multiply(A-2*E,t);

> solve({-t1-t2=1,t1+t2=-1},{t1,t2});

>

Set t1=-1 (It is the same value of t2 as in the eigenvector .), and find t1. Thus one solution of the matrix equation is . Now the second solution of the differential equation is

.