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**Unformatted text preview: **. satin 7.3 11 0 1
6'AR=(0 0);9=(1 —2) 1—4/3 —4/3 , _ —1/3 0
0 0 0 >’Q“( ' > has reduced form AR = , so rank (A) = 2. OOl—I
Ol—‘O
OWN] (b) From AR we see R1 = (1, —1,4) and R2 = (O, 1, 3) constitute a basis for the row space of
A, which is a dimension 2 subspace of R3, A 1 — 1
(c) From A R we see 01 = 0 and 02 = 1 constitute a basis for the column space
2 -1 of A, which is a dimension 2 subspace of R3. 1 —1 O O 0
involved interchanging rows 2 and 3. From AR we see rank (A)
(b) Since rows 2 and 3 were interchanged to get AR we see R1 0 1/6 1/6
0 1/6 1/6 ,which
0 6 0' O 1 1 1
4. (a) A = 12 O O 2 2 has reduced form A3 = O
O O O O
1
0
2. (6,0,0,1,1) and R3 = (1,—1,0,0,0) constitute a basis for the row space of A, which is a dimension 2 subspace of
R5.
6 O
(c) 01 = 12 ,02 = O constitute a basis for the column space of A, which is a
1 —1
dimension 2 subspace of R3.
1 0 .0 1 0 0
2 O 0 O 0 1 _
10. (a)A— 1 0 _1 reduces toAR— 0 0 0 ,sorank(A)—2.
3 O 0 0 0 O (b) The reduction required interchanging rows 2 and 3, so R1 = (1, 0,0) and R3 = (1,0,—1)
constitute a basis for the row space of A, which is a dimension 2 subspace of R3. 1 O
(c) 01 = 3 ,C2 = _01 constitute a basis for the column space of A, which is a
3 O dimension 2 subspace of R4. _ ——3 2 1 1 0 1 —2/3 —1 3 —1 3 O
14. (a) A _( 6 _4 _2 _2 0 > reduces to AR: ( O O 0/ 0/ 0 >, so
rank (A) = 1. (b) R1 = (~3,2,1, 1,0) constituta a basis for the row space of A, which is a dimension 1 Sub5f>ace 04‘ (as.
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