Math 2270, Fall 2015
Instructor: Thomas Goller
27 October 2015
Test #3 Solutions
1
Inventions (10 points)
(a) Invent a matrix A such that det A =
Possible solution: A =

2
0
4 and det(3A) =
36. (2 points)
0
2
(b) Invent a square matrix A such that rank(A) = 2 and det A = 0. (2 points)
2
3
1 0 0
Possible solution: A = 40 1 05
0 0 0
(c) Invent a vector space V whose objects are matrices such that dim V = 100. (2 points)
Possible solution: V = M4⇥25
(d) Invent a 2 ⇥ 2 matrix A such that AT 6= A and Col A = Row A. (2 points)
Possible solution: A =

1 2
3 4
2 3
1
2
4
(e) Invent a basis B = {~p1 , p~2 , p~3 } of P2 such that [t ]B = 15. (2 points)
0
Possible solution: B = {t2
1, 1, t}
Math 2270, Fall 2015
2
Instructor: Thomas Goller
27 October 2015
Subspaces Associated to a Matrix (10 points)
2
1
4
Let A = 0
2
6
0
12
3
3
0 5.
6
(a) Compute the reduced echelon form of A. (2 points)
2
1 6
Solution: 40 0
0 0
3
3
05
0
(b) Compute a basis for Nul A. What is dim(Nul A)? (4 points)
2
3
2 3
2 3
6x2 + 3x3
6
3
5 = x2 4 1 5 + x3 405, so
x2
Solution: The general solution to A~x = ~0 is ~x = 4
x3
0
1
82 3 2 39
3 =
< 6
a basis for Nul A is 4 1 5 , 405 . Thus dim(Nul A) = 2.
:
;
0
1
(c) Write down a basis for Col A. What is dim(Col A)? (2 point)
8 2 39
< 1 =
Solution: A basis for Col A is 4 0 5 , so dim(Col A) = 1.
:
;
2
(d) Write down a basis for Row A. What is dim(Row A)? (2 point)
8 2 39
< 1 =
Solution: A basis for Row A is 4 6 5 , so dim(Row A) = 1.
:
;
3
Math 2270, Fall 2015
3
Instructor: Thomas Goller
27 October 2015
Determinant (10 points)
(a) Let H be the set of all matrices A in M2⇥2 such that det A = 0. Answer each of the
following questions. You do not need to justify your answers. (4 points)
(i) Is ~0 in H?
Solution: Yes.
(ii) Is H closed under vector addition?
Solution: No.
(iii) Is H closed under scalar multiplication?
(iv) Is H a subspace of M2⇥2 ?
Solution: Yes.
Solution: No.
(b) Is the set of polynomials {1 + 2t, 3t + 3t2 , 4 t + 7t2 } a basis for P2 ? Use a coordinate
mapping to reduce the problem to computing the determinant of a 3 ⇥ 3 matrix. Show
your work! (6 points)
Solution:
Since
2 3
1
4
[1 + 2t]E = 25
0
1 0
2 3
0 3
2 3
0
2
4
[3t + 3t ]E = 35
3
4
3
1 =1
3
7
[4
2
3
4
t + 7t2 ]E = 4 15
7
1
2 3
+4
= 24 + 24 = 48,
7
0 3
the matrix is invertible, so its columns are a basis of R3 and therefore the set of polynomials is a basis of P2 .
Math 2270, Fall 2015
4
Instructor: Thomas Goller
27 October 2015
Change of Basis (10 points)
Consider the basis B =
⇢

2
2
,
1
0
of R2 .
(a) Compute the B-coordinates of each of the following vectors in R2 : (3 points)



2
2
1
(i)
Solution:
=
1
1 B
0



2
2
0
(ii)
Solution:
=
0
0 B
1



4
4
1
(iii)
Solution:
=
1 B
3
1
(b) Write down the matrix P that converts B-coordinates to standard coordinates. MulE
B
tiply P by one of your answers in (a) to check that you get back the original vector.
E
B
(3 points)
Solution:
E
P =
B


2 2
1 0
2 2
1 0

1
=1
3



2
2
4
+3
=
1
0
1
(c) Compute the matrix P that converts standard coordinates to B-coordinates. Compute

B E
2
4
P
and P
. You should get your answers in (a). (4 points)
1
B E
B E 1
Solution:
1
P = P
=
B E
E B
2



0 1
2
1
=
1
1
1
0
2
1

0
1

2
0 1
= 1
2
1
2



0 1 4
1
=
1
1 1
3
2
Math 2270, Fall 2015
5
Instructor: Thomas Goller
27 October 2015
Coordinates and Linear Transformations (10 points)
Let P2
T
/
R be the linear transformation defined by T (~p) =
2
(a) Compute T (1), T (t), and T (t2 ). (2 points)
Solution:

1
T (1) =
1
T (t) =

1
2
2

p~( 1)
.
p~(2)
T (t ) =

1
4
Let E = {1, t, t2 } be the standard basis of P2 . Using the coordinate mapping P2
S
we can view T as a linear transformation R3
/ R2
[ ]E
/
R3 ,
defined by S(~x) = A~x.
(b) Write down the matrix A. Hint: the columns of A are your answers in (a). (2 points)
Solution:
A=

1
1
1 1
2 4
(c) Use the matrix A to check that S([t2 ]E ) = T (t2 ). (1 point)
Solution:
S([t2 ]E ) =

2 3

0
1 1 4 5
1
0 =
= T (t2 )
2 4
4
1
1
1
(d) Compute a basis for Nul A. (4 points)
Solution:


1 1
1
!
2 4
0
The general solution to A~x = ~0 is
1
1

1 1
1
!
3 3
0

1 1
1 0 2
!
1 1
0 1 1
2
3
2 3
2x3
2
~x = 4 x3 5 = x3 4 15 ,
x3
1
8 2 39
< 2 =
so a basis for Nul A is 4 15 .
:
;
1
(e) Use your answer in (d) to write down a basis for the kernel of T . (1 point)
Solution: A basis for the kernel of T is { 2
t + t2 }.