Math 2270 Spring 2004
Homework 7 Solutions
Section 2.4) 14, 16, 20, 23, 25, 26, 28, 30, 35
(14) A is 2 x 2, B is 1 x 3, C is 3 x 3, D is 3 x 1 and E is 1 x 1. The following products are
defined.
"
"
# "
#
#
1 1
1 1
2 2
AA =
=
1 1
1 1
2 2
BC =
h
1 2 3
BD =
h
1 0 −1
h
i

2
1 0 
 = 14 8 2

3 2 1
1 2 3


1
h
i


6
 1  =
1
i






i


−2 −2 −2
1 0 −1
1 0 −1



 

CC =  2 1 0   2 1 0  =  4 1 −2 
10 4 −2
3 2 1
3 2 1






0
1
1 0 −1


 

CD =  2 1 0   1  =  3 

6
1
3 2 1




1 2 3
1
h
i



1 2 3 = 
DB =  1 
 1 2 3 
1 2 3
1




5
1
h
i



5 =  5 
DE =  1 

5
1
EB =
h
5
i h
EE =
(16) TRUE
h
1 2 3
5
i h
5
i
i
=
=
h
h
5 10 15
25
i
i
(In − A)(In + A) = In2 + In A − AIn − A2 = In + A − A − A2 = In − A2
(20) FALSE: This is only true for matrices that commute with one another (AB = BA).
(A − B)(A + B) = A2 + AB − BA − B 2 6= A2 − B 2
1
Math 2270 Spring 2004
Homework 7 Solutions
(23) TRUE
ABA−1
3
= ABA−1 ABA−1 ABA−1 = ABIn BIn BA−1 = AB 3 A−1
(25) TRUE: By Fact 2.4.9, A−1 B is invertible.
A−1 B
B −1 A = A−1 BB −1 A = A−1 In A = A−1 A = In
(26) Define
Then
"





1
0
0
0
0
1
0
0
A =
"
I2 I2
0 I2
#
1
0
1
0
0
1
0
1
1 2
3 4
"
 








#
A B
0 A
1
3
0
0
2
4
0
0
B =
#
2
4
1
3
=
3
5
2
4
"





"
2 3
4 5
#
#
A B+A
0
A




= 
1
3
0
0
2
4
0
0
3
7
1
3
5
9
2
4
(28) We look at a general 2 x 2 matrix A with real entries.
"
a b
c d
#
"
a b
c d
#
=
"
a2 + bc ab + bd
ac + cd bc + d2
#
=
"





0 0
0 0
#
The final matrix has four entries, giving us a system of four equations and four unknowns.
a2 + bc = 0
ab + bd = b(a + d) = 0
ac + cd = c(a + d) = 0
bc + d2 = 0
The second and third equations tell us that either b = 0 and c = 0, or a + d = 0. If b
and c are both zero, then a and d must also be zero for the first and fourth equations to
be true. This gives the zero matrix. To find a nonzero matrix, we assume that b and c are
nonzero and a + d = 0, or a = −d. Then, from the first and fourth equations we see that
2
Math 2270 Spring 2004
Homework 7 Solutions
a2 = −bc = d2 must be true. One way to satisfy this (but not the only way) is to let b = a,
c = −a, and d = −a. If a = 1 we have
"
1 1
−1 −1
#
(30) If A is a noninvertible n x n matrix, we can always find a nonzero n x n matrix B
such that AB = 0. We represent B by its column vectors v~1 , ... v~n . Then, the product
AB = A [~v1 ~v2 · · · ~vn ] = [A~v1 A~v2 · · · A~vn ]. We can think of the equation AB = 0 as the
system of equations A~v1 = 0, A~v2 = 0, ..., A~vn = 0. A is noninvertible, so there are infinitely
many solutions to the system A~x = ~0 by Fact 2.3.4. Therefore, we can find nonzero vectors
~v1 , ~v2 , ..., ~vn with which to construct the desired matrix B.
(35) Assume A is n x m, B is m x n and BA = Im .
(a) If ~x is a solution of the system A~x = ~0, then it must also satisfy BA~x = B~0. But,
BA = Im and B~0 = ~0, so we must have Im~x = ~x = 0. Therefore, the only solution to the
system A~x = ~0 is the zero vector.
(b) For all vectors ~b ∈ ℜm , the linear system B~x = ~b always has the solution ~x = A~b. To see
this, multiply B~x = B(A~b) = (BA) ~b = Im~b = ~b.
(c) From part (a), we see that there must be a leading one in every column of rref(A).
Therefore rank(A)= m. From part (b), the system B~x = ~b is consistent for all ~b ∈ ℜm , so
rref(B) cannot have a row of zeros. Therefore, rank(B)= m.
(d) From part (c), rank(A)=rank(B)= m. In example 3 in section 1.3 (pg. 25) we saw that
if a matrix is m x n, then its rank must be less than or equal to both m and n. Therefore,
m ≤ n.
Section 3.1) 2, 4, 16, 17, 18, 32, 34
(2) We must solve the system A~x = ~0.
"
2 3 | 0
6 9 | 0
#
→
"
1 3/2 | 0
0 0 | 0
#
The rref(A) tells us that there are infinitely many solutions of the form
~x =
"
−3/2 t
t
#
= t
"
−3/2
1
#
=
1
t
2
"
−3
2
#
where t is any real number. Therefore, the kernel of A is spanned by the vector
3
"
#
−3
.
2
Math 2270 Spring 2004
Homework 7 Solutions
(4) We must solve the system A~x = ~0 where A = [1
Solutions are of the form


−2

~x = r 
 1  + s
0
 

2 3]. Both ~x2 and ~x3 are free variables.


−3


 0 
1

−3
−2

 

Thus ker(A)=span 1  ,  0 .
1
0
(16)








1
3
2
1








A~x = x1  1  + x2  2  + x3  3  = (x1 + 2x2 + 3x3 )  1 
1
3
2
1


1


im(A) = span  1 

1
(17) The image of A is the x1 -x2 plane in ℜ2 , or all of ℜ2 .
"
im(A) = span
1
3
# "
,
1
2
#!
(18) We notice that the second column of A is four times the first column of A. Therefore,
im(A) = span
"
1
3
#!
which is a line in ℜ2 .
(32) Take T to be the linear transformation defined by T (~x) = A~x where


7


A= 6 
5
4
Math 2270 Spring 2004
Homework 7 Solutions


−1


(34) We must find a matrix A such that A  1  = ~0. Think of the matrix A in terms of
2




w
~1
−1

 w
~2 

. Then, the vectors w
~ i must satisfy w
~i · 

its row vectors A = 
 1  = 0. In other
 ··· 
2
w
~n
words, the w
~ i vectors must be orthogonal (perpendicular) to the given vector spanning the
kernel. Two possibilities are w
~ 1 = [1 1 0] and w
~ 2 = [2 0 1]. A possible transformation T
is that defined by T (~x) = A~x where
A=
"
1 1 0
2 0 1
#
Notice that the transformation matrix must have at least two rows. If we form a 1 x 3 matrix
with only one of these vectors, say w
~1 , we could find vectors in the kernel
of A that are not

−1
−1




scalar multiples of the vector  1 . For example, the vector  1  would also be in the
0
2
kernel of A.
5