Example 2: Solve the system .

Solution:

> A:=matrix(3,3,[0,1,1,1,0,1,1,1,0]);

> E:=array(1..3,1..3,identity);

> t:=matrix(3,1,[t1,t2,t3]);

> eigenvals(A);

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

We know that the eigenvalue 2 has only one eigenvector. We set t1=t2=1 and we find t3 as follows

> solve(-2+1+t3=0,t3);

Hence one eigenvector is and the corresponding solution is

. On the other hand eigenvalue -1 has two independent eigenvectors.

> multiply(A+E,t);

> solve({t1+t2+t3=0,t1+t2+t3=0,t1+t2+t3=0},{t1,t2,t3});

Hence there are two independetn eigenvectors which we can obtain by setting first t2=0 and t3=-1, and tehn by setting t2=1 and t3=-1..

Hence two independent eigenvectors that correspond to -1 are and and thus two independent solutions of the differntial equation are