F2010 M341 Homework 5
Section 2.4.
2.b)
3

1
4
2

2
1

1
3
2
∼
1

3

2
0
4
5
0
8
10
∼
1

3

2
0
4
5
0
0
0
. The rank of this matrix is 2
(number of non all zero rows). The matrix is singular (not invertible).
3.b)
The matrix
(
10

5

4
2
)
is singular (
ad

bc
= 10
×
2

(

5)(

4) = 0).
Other (better) method: By the row operation
<
2
>
←
<
2
>
+4
/
10
<
1
>
we have:
(
10

5

4
2
)
∼
(
10

5
0
0
)
. So the rank is only 1 which means that the matrix is singular.
4.b)
5
7

6

1
0
0

3
1

2

0
1
0
1

5
2

0
0
1
∼
1
7
/
5

6
/
5

1
/
5
0
0

3
1

2

0
1
0
1

5
2

0
0
1
∼
1
7
/
5

6
/
5

1
/
5
0
0
0

16
/
5

8
/
5
 
3
/
5
1
0
0

32
/
5
16
/
5
 
1
/
5
0
1
∼
1
7
/
5

6
/
5

1
/
5
0
0
0
1

1
/
2

3
/
16

5
/
16
0
0

32
/
5
16
/
5
 
1
/
5
0
1
∼
1
7
/
5

6
/
5

1
/
5
0
0
0
1

1
/
2

3
/
16

5
/
16
0
0
0
0

1

2
1
Unless I did a mistake (always possible with
these heavy computations, the rank is 2, less than 3 and the matrix is singular.
7 b)
the matrix of coeﬃcients of the system is
A
=

5
3
6
3

1

7

2
1
2
.
By row reducing [
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.
 Spring '08
 HIETMANN
 Diagonal matrix, Triangular matrix

Click to edit the document details