Math 2270, Fall 2014
Instructor: Thomas Goller
30 September 2014
Test #2 Solutions
1
Elimination (12 points)
Find the complete solution to the
2
2
4
2
A=
0
Show your work!
equation A~x = ~b, where
3
2 3
2 1 4
1
~
5
4
2 0
5
and
b = 25 .
0 2
2
6
h
i
Solution: Elimination on A ~b yields
2
3
2
2 2 1 4 1
2 2
h
i
4
5
4
~
1 3
! 0 0
A b ! 0 0 1
0 0 2
2 6
0 0
2
2 2
4
! 0 0
0 0
3
4 1
⇥
⇤
1 35 = U ~c !
0 0
3
2
3
0 5
2
1 1 0 52
1
h
i
5
4
5
~
1 3 = R d .
1
1 3
! 0 0 1
0 0
0
0 0 0 0
0
⇥
⇤
(We could also have stopped elimination at U ~c .)
The free columns in R are columns 2 and 4. To find the particular solution ~xp , we solve
2 3
2
3 x1
2 3
5
1 1 0 2 6 7
1
07 4 5
40 0 1
15 6
3
=
,
4 x3 5
0 0 0 0
0
0
2 3
1
607
7
which by back substitution yields ~xp = 6
4 3 5. To find the special solutions that span the
0
nullspace N (A), we need to solve
2 3
2 3
2
3 x1
2 3
2
3 x1
2 3
5
5
1 1 0 2 6 7
0
1 1 0 2 6 7
0
17 4 5
07 4 5
6
6
40 0 1
5
4
5
1 4 5= 0
0 0 1
1 4 5= 0 ,
and
x3
x3
0 0 0 0
0
0 0 0 0
0
0
1
2 3
2 53
1
2
617
6 0 7
~
7
6 7
which yield ~s1 = 6
4 0 5 and ~s2 = 4 1 5. Thus the complete solution to A~x = b is
0
1
2 3
2 53
2 3
1
1
2
607
617
6 0 7
7
6 7
6 7
~x = 6
4 3 5 + x2 4 0 5 + x4 4 1 5 .
0
0
1
1
1
0
Math 2270, Fall 2014
2
Instructor: Thomas Goller
30 September 2014
Inventions (10 points)
(a) Invent a 2⇥2 matrix that is not invertible and whose entries are all non-zero. (2 points)
Possible solution:

1 1
1 1
(b) Invent a 2 ⇥ 3 row-reduced echelon matrix R whose nullspace is spanned by one non-zero
vector. (2 points)
Possible solution:
R=

1 2 0
0 0 1
(c) Invent 2 ⇥ 2 matrices A and B such that (AB)T 6= AT B T . (2 points)
Possible solution:

1 0
A=
,
2 3
B=

0 2
0 1
(d) Invent a matrix A whose column space is R3 . (2 points)
Possible solution:
2
3
1 0 0 0
A = 40 0 1 05
0 0 0 1
T
(e) Invent vectors ~v and w
~ so that ~v w
~ =
Possible solution:


2 3
. (2 points)
4 6
1
~v =
,
2
w
~=

2
.
3
Math 2270, Fall 2014
3
Instructor: Thomas Goller
30 September 2014
Rank (9 points)
(a) Define the rank r of a matrix A. (2 points)
Solution:
The rank r of A is the number of pivots of A.
(b) Suppose A is a 5 ⇥ 3 matrix. What are the possibilities for r? (2 points)
Solution:
r could be 0, 1, 2, or 3.
(c) Invent a 5⇥3 matrix A with r = 3. Explain why the nullspace N (A) is the zero subspace
{~0}. (3 points)
Possible solution:
2
1
60
6
A=6
60
41
0
0
1
0
2
4
3
0
07
7
17
7
35
7
Since every column is a pivot column, there are no free columns, so there are no special
solutions. Since N (A) is the span of the special solutions, N (A) must be {~0}.
(d) Using N (A) = {~0} as in part (c), what can you deduce about the number of solutions
of A~x = ~b? (2 points)
Solution: For each ~b, the equation A~x = ~b has either no solution or only one solution.
Math 2270, Fall 2014
4
Instructor: Thomas Goller
30 September 2014
Subspaces (9 points)
Recall that a subset S of a vector space is called a subspace if two conditions hold:
(i) For every vector ~v in S, every c~v is still in S.
(ii) For every two vectors ~v and w
~ in S, the sum ~v + w
~ is still in S.
For each S defined below, decide whether condition (i) holds, whether condition (ii) holds,
and whether S is a subspace. (You do not need to show any work.)
(a) The line S in R2 with equation 2x
3y = 4. (3 points)
Solution: (i) does not hold, (ii) does not hold, S is not a subspace
02 3 2 3 1
1
0
@
4
5
4
1 , 15A in R3 . (3 points)
(b) The plane S = span
1
0
Solution: (i) holds, (ii) holds, S is a subspace
(c) The first quadrant S in R (the set of all vectors
2

x
with x
y
Solution: (i) does not hold, (ii) holds, S is not a subspace
0 and y
0). (3 points)
Math 2270, Fall 2014
5
Instructor: Thomas Goller
30 September 2014
Inverses (10 points)
(a) Suppose A is an invertible n ⇥ n matrix and ~b is in Rn . What is the unique solution ~x
to A~x = ~b? (2 points)
Solution:
~x = A 1~b
(b) Suppose A is invertible and that B and C are two inverses of A. Show that B = C.
(This proves that an invertible matrix has a unique inverse!) Hint: use associativity on
the product BAC and the definition of inverse. (4 points)
Solution:
B = BI = B(AC) = (BA)C = IC = C.
(c) What is the inverse E211 of the elimination matrix E21
Solution:
E211
(d) What is the inverse P
1
2
3
1 0 0
= 4 2 1 05? (2 points)
0 0 1
2
3
1 0 0
= 4 2 1 05
0 0 1
2
3
0 1 0
of the permutation matrix P = 40 0 15? (2 points)
1 0 0
Solution:
P
1
2
3
0 0 1
= 4 1 0 05
0 1 0