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AME525_Homework_04_Solutions_100307

# AME525_Homework_04_Solutions_100307 - satin 7.3 11 0 1...

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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. (Q g! = <1; > @th“ a. (0051s ear Hz Co‘umn great: 04‘ {.3 t which 1‘; o. ; Madam ‘60. mag R. Gauss-Jordan Problems q) {1+2lesz =1 (9 2, 021 5445-2063 211*27‘2’37‘3 - 5 @ x1— «2 7316 . 5 @ 2(1-2xf2x3) +2xzfsf>c3 - J ’ . 212*603 a—i @ gwxzwij- (1-2912‘2063) * x2 r2295, = 3 “”2" *3 ° 4 @ 405-1)ij =4 2} 435-41% ..__ 2 2 \ Mm “awaég «a a - Milka «2% .IMLQ'} «1 a yawn-2(1) = 1 :1? ...
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AME525_Homework_04_Solutions_100307 - satin 7.3 11 0 1...

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