This preview shows pages 1–2. Sign up to view the full content.
Prelim 1
Short solutions (corrected)
September 28, 2006
1.
(a)
x
1
x
2
x
3
x
4
x
5
x
6
=
1
0
2
0
0
3
+
r

3
1
0
0
0
0
+
s
1
0

1
1
0
0
+
t

1
0

2
0
1
0
;
r, s, t
∈
R
(b)
p
= 6
, q
= 4
(c) rank(
A
) = dim (image(
T
)) = 3
(d) Yes, because
A~x
=
~
b
has a solution.
(e) dim (ker(
T
)) = 3, basis is the last 3 vectors in (a).
(f) Basis for image(
T
) is
{
v
1
, v
3
, v
6
}
; not enough information to de
termine them explicitly.
2.
(a)
A

1
=

3

2
4
2
1

2

1
0
1
; check
AA

1
=
I
3
.
(b)
~v
=
A

1
1
1
1
=

1
1
0
3.
(a) For example,
v
1
=
1

1
2
,
v
2
=
1
1
0
,
v
3
=
0
2
1
.
(b)

1 0 0
0
1 0
0
0 1
(c)
1
1 0

This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
This note was uploaded on 10/01/2008 for the course MATH 2940 taught by Professor Hui during the Fall '05 term at Cornell University (Engineering School).
 Fall '05
 HUI
 Linear Algebra, Algebra

Click to edit the document details