1. Problem 3 on Page 151.
Solution
.
0 0 1
0 1 0
1 0 0

1
=
0 0 1
0 1 0
1 0 0
,
1 0 0
0 3 0
0 0 1

1
=
1
0
0
0 1
/
3 0
0
0
1
,
1
0 0
0
1 0

2 0 1

1
=
1 0 0
0 1 0
2 0 1
.
2. (i) Find the rank of the matrices in Problem 2 (b), (f) on Page 165.
(ii) For the matrices in Problem 2 (b), (f) on Page 165, ﬁnd a basis for its
row space (the vector space spanned by the row vectors of the matrix) and a
basis for its cloumn space (the vector space spanned by the column vectors
of the matrix),
Solution
. (i) 2 (b)
1 1 0
2 1 1
1 1 1
→
1
1
0
0

1 1
0
0
1
→
1 0 0
0 1 0
0 0 1
.
So its rank is 3 (i.e. the matrix is invertible).
2 (f)
1
2
0
1
1
2
4
1
3
0
3
6
2
5
1

4

8 1

3 1
→
1 2 0 1
1
0 0 1 1

2
0 0 2 2

2
0 0 1 1
5
→
1 2 0 1
1
0 0 1 1

2
0 0 0 0
2
0 0 0 0
7
→
1 2 0 1 0
0 0 1 1 0
0 0 0 0 1
0 0 0 0 0
So its rank is 3.
1
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 Summer '08
 Staff
 Linear Algebra, Algebra, Matrices, Row echelon form, Solutition.

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