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.
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.
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.
Answer This subset of does not contain the zero vector. Hence, it is not a subspace.
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
, , , , , , , 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 , , , , , , , .
First Method: where M is the matrix in part (a).
Second Method: Calculate directly the determinant.
. Hence the volume is 3|c|.
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:
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 .
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:
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,