Homework #11

Section 5.3. Ex. 6a)  If the vectors were on the same line they would be all multiples of the same vector. Since this is not the case, the vectors do not belong to the same line.

  Ex. 8)  We can see that v1+v2-v3=0 so the three vectors are linearly dependent. From that equation we can solve for v3, v2 and v1 in terms of the other vectors. For example v3=v1+v2, v2=v3-v1, v1=v3-v2.

Ex 9) . The vectors form a lineary dependent set iff the matrix formed with their components has less than three leading ones. This happens if the determinant of the matrix if zero. So we will solve for the determinant:

>	with(LinearAlgebra):

>	A:=Matrix([[lambda,-1/2,-1/2],[-1/2,lambda, -1/2],[-1/2,-1/2,lambda]]);

>	d:=Determinant(A);

>	solve(d=0,lambda);

As it can be seen, the solutions to the determinant of the matrix equal zero are 1, and -1/2.

Section 5.4 Ex. 4b)  Notice that the sum of the second and third polynomials equals the first polynomial, therefore the three vectors are not linearly independent. Since the dimension of P2 is 3, we need three vectors for a basis, therefore the given vectors cannot be a basis in P2.

Ex. 7b)  We need to solve the vector equation: c1u1+c2u2=w. This leads to a linear system whose augmented matris is:

>	A1:=Matrix([[2,3,1],[-4,8,1]]);

>	ReducedRowEchelonForm(A1);

The values in the third column in the RREF above are the coeeficients c1 and c2 respectively.

Ex. 9a)  As above, we need to solve the vector equation: c1v1+c2v2+c3v3=v, whihc leads to a system

>	A2:=Matrix([[1,2,3,2],[0,2,3,-1],[0,0,3,3]]);

>	ReducedRowEchelonForm(A2);

The solution is seen in the fourth column in the RREF above.

Ex. 14) . We write the matrix of the system and reduce it:

>	A3:=Matrix([[1,-4,3,-1],[2,-8,6,-2]]);

>	ReducedRowEchelonForm(A3);

The solution set is given by: (4u-3s+t,u,s,t)=u(4,1,0,0)+s(-3,0,1,0)+t(1,0,0,1). So the dimension of the solution space (which is the nullspace of the matrix) is three, and the basis vectors appear above. They can be obtained with Maple as well:

>	NullSpace(A3);

Ex. 18a) Look at the given equation as a system in three unknowns with only one equation. We have to find the nullspace of the matrix again. With z=3t, y=3s we get x=2s-5t, so the solution is: (2s-5t,3s,3t)=s(2,3,0)+t(-5,0,3), and again the two vectors are a basis in the solution space. The dimension is 2.

18c)  We have a line, and if we write it in vector form: (2t, -t, 4t)=t(2,-1,4), we see that a basis is (2,-1,4).

Section 5.5 Ex. 3c)  Let us write the augmented matrix. The vector b is in the column space of the matrix if and only if the augmented matrix does not have a leading one in the last column, which ie equivalent to saying that rank(A4)=rank(A4|b) (A4 is the name of the matrix in the Maple output).

>	A4:=Matrix([[1,-1,1,5],[9,3,1,1],[1,1,1,-1]]);

>	ReducedRowEchelonForm(A4);

As you can see rank(A4)=3 so b is in the column space of the given matrix.

Ex. 6c).  We let Maple do everything (we trust it, don't we?)

>	A5:=Matrix([[1,4,5,2],[2,1,3,0],[-1,3,2,2]]);

>	ReducedRowEchelonForm(A5);

>	NullSpace(A5);

Ex. 11b) We form a matrix in which the given vectors are rows. Then we will find a basis for the Row space of the amatrix by getting the RREF of the matrix.

>	A6:=Matrix([[-1,1,-2,0],[3,3,6,0],[9,0,0,3]]);

>	ReducedRowEchelonForm(A6);

Since we get three leading ones, the three vectors are linearly independent, so a basis the space they span can be formed with the original vectors, but the rows in the above RREF can be chosen as well. Here is Maple at work:

>	RowSpace(A6);