Math 235 Solution of Midterm 2 Spring 1999
a) (9 points) Find a basis for the null space Null(A) of A.
Recall that Null(A) is the space of solutions of the linear system . Since A and B are row equivalent, Null(A)=Null(B). We have three free variables: , , and . Hence Null(B) is 3 dimensional. The equations
and the three vectors are a basis of Null(A).
b) (9 points) Find a basis for the column space of A.
Take the pivot columns of A :
c) (6 points) Is the vector in the column space of A? Justify your answer!
No! The vector b is in Col(A) if and only if the system is consistent. We could check that it is inconsistent by row reduction of the augmented matrix. However, this would require a lot of work. Instead, use the fact that b is in Col(A) if and only if the vector b is a linear combination of the basis vectors in part (b). I.e., if and only if the augmented matrix does not have a pivot in the fourth column. But it does (Check by row reduction).
Regard the equation of the plane as a linear ``system'' of one equation. So y and z are free variables and the general solution has the form
(a) (9 points)
This is a subspace.
It is the set of vectors of the form Hence, it is the subspace spanned by these four vectors. Equivalently, it is the column space of the matrix having these 4 vectors as column vectors.
(b) (9 points)
This is not a subspace because the zero vector is not in this set (the equation is not homogeneous).
(c) (6 points)
This is a subspace. If we rewrite the equations in the form we get that this set is Null(A) where .
Translating by the vector (-1,-1) we get the parallelogram with vertices (0,0), (2,3), 5,1), (7,4) whose area is
b) (6 points) Compute the volume of the parallelepiped in with vertices
, , , , , , , where
The volume is the absolute value of the following determinants
c) (4 points)
Use your answer in part (b)
and the algebraic properties of determinants
to compute the volume of the
parallelepiped obtained if is replaced by
where a, b, c are real numbers. (Express your answer in terms of a, b, c).
By the algebraic properties of determinants we know that
The first and second equalities follow from the fact that adding a multiple of one column to another does not change the determinant. The third equality reflects the property that if a matrix B is obtained from A by multiplication of a column by a scalar c then . The last equality is our computation in part (b). A more computational method is to express in terms of a, b, c
with A invertible. Solve for C in terms of A and B.
b) (8 points) The inverse of is .
c) (4 points) Compute the (2,2) entry of C in part (a) if A is given in part (b) and
. The second row of is (2, 1, 0). Multiplying by the second column of A we get .
(a) (6 points) Show that the subset H of of polynomials p(t) of degree which in addition satisfy
is a subspace of . (The straightforward answer would include the definition of a subspace and a verification that H satisfies all the properties.)
Answer: A polynomial p(t) of the form is in H if and only if
So H is the subset of of polynomials with the property that the sum of their coefficients is 0. To show that it is a subspace you can identify with by mapping the polynomial to the vector and argue that (1) is the equation of a plane in through the origin. Alternatively, you can argue more abstractly using the definition of a subspace of a vector space:
1) The zero polynomial is in H.
2) If and are in H, then p+q is in H because (p+q)(1)=p(1)+q(1)=0+0=0.
3) If p is a polynomial in H and c is a scalar then cp is in H because .
(b) (4 points) Find a basis for H. Explain why the set you found is linearly independent and why it spans H.
Answer: As a basis to the plane in given by equation (1) we can take the two vectors (use the same method as in problem 2). Thus, the two polynomials
are a basis for H.