THE CHINESE UNIVERSITY OF HONG KONG
Department of Mathematics
MAT2310B Linear Algebra and Applications (Fall 2007)
Midterm Examination Solutions
1. (a) First we perform the elementary row operation to reduce the matrix
A
.
•
(

35
/
2) times the first row and add to the second row
•
(

3) times the first row and add to the third row,
we obtain
2

1
2
0
31
/
2

39
0

1

5
so
m
21
= 35
/
2
, m
31
= 3.
•
(2
/
31) times the second row and add to the third row,
we obtain
2

1
2
0
31
/
2

39
0
0

233
/
31
so
m
32
=

2
/
31.
Hence
U
=
2

1
2
0
31
/
2

39
0
0

233
/
31
L
=
1
0
0
35
/
2
1
0
3

2
/
31
1
We have gain the decomposition
A
=
LU
(b) Now solving
A
x
=
b
is equivalent to solving
LU
x
=
b
. We can solve
LU
x
=
b
by the
following two steps. First, let
U
x
=
z
and then solve
L
z
=
b
by forward substitution. We
obtain
z
1
=
1
z
2
=
4

(35
/
2)
z
1
=

27
/
2
z
3
=
3

(3)
z
1
+ (2
/
31)
z
2
=

27
/
31
Next we solve
U
x
=
z
and get
x
3
=
(

31
/
233)
×
(

27
/
31) = 27
/
233
x
2
=
2
31
((

27
/
2) + 39
x
3
) =

135
/
233
x
1
=
1
2
(1 +
x
2

2
x
3
) = 22
/
233
Thus the solution to the given linear system is
x
=
22
/
233

135
/
233
27
/
233
.
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2
2. (a) Let
A
be
1
1
α
1
α
1
α
1
1
The system has an unique solution if and only if det(
A
)
6
= 0
But det(
A
) =
α
+
α
+
α

α
3

1

1 =

α
3
+ 3
α

2
which equals to

(
α

1)
2
(
α
+ 2)
Hence the system has an unique solution if and only if
α
6
= 1 and
α
6
=

2
(b) The augmented matrix of the given system is
1
1
α
α
2
1
α
1
α
α
1
1
1
•
(1) times the first row and add to the third row
•
(

α
) times the second row and add to the third row,
we obtain
1
1
α
α
2
0
α

1
1

α
α

α
2
0
1

α
1

α
2
1

α
3
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 Spring '06
 Dr.JeffChakFuWONG
 Linear Algebra, Algebra, Row, 0 K, Elementary matrix, 0 span

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