Example 5: Solve the system .

Solution:

> A:=matrix(3,3,[5,-3,-2,8,-5,-4,-4,3,3]);

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

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

> eigenvalues(A);

Here, 1 one is a triple eigenvalue.

> multiply(A-1*E,t);

> solve({4*t1-3*t2-2*t3=0,8*t1-6*t2-4*t3=0,-4*t1+3*t2+2*t3=0},{t1,t2,t3});

The eigenvalue 1 has two independent eigenvectors, and we could obtain them by setting first t1=1 and t2=0, and then by setting t1=0 and t2=1.

Hence two independent egienvectors corresponding to the eigenvalue 1 are and ., and the corresponding two independent solutions of the differential equation are and

. We find a third independent solution in the similar way as we did it in Example 4. We have to solve the matrix equation . This system does not have a solution for all choices of and . The second row of is a multiply of the first row, so on the right hand side of the equation we should also have that the second row be twice the first, that is the system will have a solution only if . Now set .

> solve({4*t1-3*t2-2*t3=1,8*t1-6*t2-4*t3=2,-4*t1+3*t2+2*t3=-1},{t1,t2,t3});

Thus one solution of thes one solution of the matrix equation that corresponds to and is obtained by setting t1=0 and t2=0 to get t3=-1/2. Hence a third independent solution of the differential equation is that is .