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Solutions to problems from section 1.7
2. Determine if the vectors
0
0
2
,
0
5

8
,

3
4
1
are linearly independent.
Solution:
Consider the matrix
A
whose columns are the vectors listed above. The vectors
above are linearly independent if and only if the equation
A
x
=
0
has only the trivial solution,
which happens if and only if the matrix
A
has a pivot position in each column (otherwise
there would be free variables). Well,
A
=
0
0

3
0
5
4
2

8
1
R
1
↔
R
3
→
2

8
1
0
5
4
0
0
3
has a pivot position in each column, thus the vectors
0
0
2
,
0
5

8
,

3
4
1
are linearly
independent.
6. Determine if the columns of the matrix
4

3 0
0

12 4
1
0 3
5
4 6
are linearly independent.
Solution:
Using the same argument as in exercise 2 we just need to check whether or not
the matrix above has a pivot position in each column:
4

3 0
0

12 4
1
0 3
5
4 6
R
1
↔
R
3
→
1
0 3
0

12 4
4

3 0
5
4 6
NR
3
=
R
3

4
R
1
NR
4
=
R
4

5
R
1
→
1
0
3
0

12
4
0

3

12
0
4

6
R
2
↔
R
3
→
1
0
3
0

3

12
0

12
4
0
4

6
NR
2
=

1
3
R
3
→
1
0
3
0
1
4
0

12
4
0
4

6
NR
3
=
R
3
+12
R
2
NR
4
=
R
4

4
R
2
→
1 0
3
0 1
4
0 0
16
0 0

22
NR
4
=
R
4
+
22
16
R
3
→
1
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 Fall '08
 AXENOVICH
 Vectors

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