This preview shows page 1. Sign up to view the full content.
(f)
1
2
3
4
5
6
5
7
9
R
2
→
R
2

4
R
1
R
3
→
R
3

5
R
1
1
2
3
0

3

6
0

3

6
R
3
→
R
3

R
2
1
2
3
0

3

6
0
0
0
.
Hence
A
is singular by virtue of the zero row.
15. Suppose that
A
is non–singular. Then
AA

1
=
I
n
=
A

1
A.
Taking transposes throughout gives
(
AA

1
)
t
=
I
t
n
= (
A

1
A
)
t
(
A

1
)
t
A
t
=
I
n
=
A
t
(
A

1
)
t
,
so
A
t
is non–singular and (
A
t
)

1
= (
A

1
)
t
.
16. Let
A
=
•
a
b
c
d
‚
, where
ad

bc
= 0. Then the equation
A
2

(
a
+
d
)
A
+ (
ad

bc
)
I
2
= 0
reduces to
A
2

(
a
+
d
)
A
= 0 and hence
A
2
= (
a
+
d
)
A
. From the last
equation, if
A

1
exists, we deduce that
A
= (
a
+
d
)
I
2
, or
•
a
b
c
d
‚
=
•
a
+
d
0
0
a
+
d
‚
.
Hence
a
=
a
+
d, b
= 0
, c
= 0
, d
=
a
+
d
and
a
=
b
=
c
=
d
= 0, which
contradicts the assumption that
This is the end of the preview. Sign up
to
access the rest of the document.
Unformatted text preview: A is non–singular. 17. A = 1 a ba 1 cbc 1 R 2 → R 2 + aR 1 R 3 → R 3 + bR 1 1 a b 1 + a 2 c + ab abc 1 + b 2 R 2 → 1 1+ a 2 R 2 1 a b 1 c + ab 1+ a 2 abc 1 + b 2 R 3 → R 3( abc ) R 2 1 a b 1 c + ab 1+ a 2 1 + b 2 + ( cab )( c + ab ) 1+ a 2 = B. 24...
View
Full
Document
This note was uploaded on 12/19/2011 for the course MAS 3105 taught by Professor Dreibelbis during the Fall '10 term at UNF.
 Fall '10
 Dreibelbis

Click to edit the document details