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Applied Linear Algebra, MATH 2702010 Fall
Midterm Quiz: Solutions
(October 20, 2010.)
1. (25 points)
Given the following matrix:
A
=

1
3
2
0
1
1
2

2
0
1

3

2
⇒
1

3

2
0
1
1
0
0
0
0
0
0
(a) rank(
A
) = 2;
(b) null(
A
) = 1; and the basis of ker(
A
) = (1
,
1
,

1)
T
.
(c) Let
b
= (5
,
2
,c,d
)
T
. Find the values of
c,d
, such that the equation
A
x
=
b
has a solution. The augmented matrix can reduced to the following
echelon form:
A
#
=

1
3
2
5
0
1
1
2
2

2
0
c
1

3

2
d
⇒
1

3

2
5
0
1
1
2
0
0
0
2 +
c
0
0
0
5 +
d
The consistency condition yields
c
=

2
,d
=

5.
2. Given the following elements in the vector space (
M
2
,
3
):
A
=
•
1 2 3
4 0 5
‚
;
B
=
•
2
4 7
10 1 13
‚
;
C
=
•
1 2 5
8 2 11
‚
.
(a) The usual basis (
B
) of the space (
M
2
,
3
) is:
E
1
=
•
1 0 0
0 0 0
‚
;
E
2
=
•
0 1 0
0 0 0
‚
;
E
3
=
•
0 0 1
0 0 0
‚
;
E
4
=
•
0 0 0
1 0 0
‚
;
E
5
=
•
0 0 0
0 1 0
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