Math 235 Section 1 Solution of Midterm 2 Fall 2000
a) (10 points) Find a spanning set for (i.e., a set of vectors which spans) the null space Null(A) of A.
Answer:
We use the row reduced echelon form to write, in parametric form,
the general solution of 
(i.e., the general vector in Null(A)).
The free variables are: 
, 
, and 
.
Hence, the general solution is
and the three column vectors above span Null(A).
b) (8 points) Are the column spaces 
and 
equal? Justify your answer carefully!
(either explain why they are equal, or
find a vector in one that does not belong to the other).
Answer: No, the two column spaces are different! For example, the last column of A has a non-zero fourth entry, while any vector in Col(B) is a linear combination of the columns of B, and hence, its fourth entry is zero.
be the 
identity matrix and
.
a) (5 points) Show that the matrix A-5I is invertible.
Note that A-5I is simply
.
Answer
. Since 
, the matrix A-5I is invertible.
b) (5 points) Express the determinant of the matrix A-tI in terms of the
scalar parameter t.
Answer
.
c) (6 points) Show that there are precisely two values of t for which the matrix A-tI is not invertible. Find these values of t.
Answer
A-tI is not invertible if and only if
its determinant vanishes.
We find those values of t buy solving the quadratic equation
. The two solutions are t=6 and t=-1.
is a subspace.
If it is not, find a property in the definition of a subspace which this set
violates.
If it is a subspace, find a matrix A such that this set is either
Null(A) or Col(A).
Answer This subset of 
does not contain the zero vector.
Hence, it is not a subspace.
Answer
We can write the general element as a linear combination
Hence, this subset is a subspace, which is equal to the
column space of the matrix
Answer:
This subset of 
is the general solution of the system
of homogeneous linear equations
It is hence a subspace which is equal to the Null space of
with vertices
, 
, 
, 
, 
, 
, 
, 
where
Answer:
The volume is the absolute value of the determinant of the 
matrix M with columns 
and .
b) (8 points) Let T be the linear transformation from to
sending a vector 
to 
, where
A is the matrix
Compute the volume of the
parallelepiped obtained as the image of the one in part (a)
under the transformation T.
Express your answer in terms of a, b and c.
Note: The image is the parallelepiped with vertices
, 
, 
, 
, 
, 
, 
,
.
Answer:
First Method:
where M is the matrix in part (a).
Second Method: Calculate directly the determinant.
.
Hence the volume is 3|c|.
matrices
satisfying the equation
with A invertible. Solve for C in terms of A and B.
Answer: Multiply both sides by 
on the left to get
. Then multiply both sides by A on the right
to get 
.
b) (10 points) Let 
Compute its inverse . (Check that 
.)
Answer: Augment A by the identity matrix and row reduce:
Hence, 
c) (6 points)
Compute the (2,3) entry of 
if A is given in part (b) and
Answer: Calculate the product of the second row of A times B. Then
take the product of the resulting row vector with the third column of
.
be the vector space of polynomials of degree 
.
Recall that a vector in is a polynomial p(t) of the form
where the coefficients 
are arbitrary real numbers.
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:
We check the three conditions for a subset of the vector space
to be a subspace:
i) The zero vector 
has value 0 when t=0 and when t=1.
Hence, the zero vector is in H.
ii) (H is closed under addition) If p and q are in H, then they satisfy equation (1). Consequently, p(0)=0, p(1)=0, q(0)=0, and q(1)=0. We need to check that their sum satisfied equation (1) as well. Indeed:
(p+q)(0)=p(0)+q(0)=0+0=0 and (p+q)(1)=p(1)+q(1)=0+0=0.
iii) (H is closed under scalar multiplication) We need to check that if p is in H and c is a scalar, then cp satisfy equation (1) as well. Indeed:
and
.
b) (4 points) Find a spanning set for H. Explain why the set you found spans H.
Answer: We can rewrite equation (1) explicitly in the form
We find the general solution,
regarding 
, 
and 
as unknowns:
is free and
If we choose 
, we get the polynomial 
.
Any other polynomial in H is a scalar multiple of this one.
Hence,
