AME525_Homework_05_Solutions_101007

# AME525_Homework_05_Solutions_101007 - 6 —1 1 O O 10 O O...

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Unformatted text preview: 6 —1 1 O O 10 O O 6. The coefﬁcient matrixA= 1 O O —1 2 has reducedformAR= O 1 —1 O 1 0 O O —2 O O O O 2 1 12 With x3=a,x5=ﬁthe general solution can be writtenasX=a 1 +ﬁ 0 so the O 4 O 1 solution space of AX =Ohas dimension2=m— rank (A)=5—3. 8 O —2 O O 1 . . 2 —1 O 3 O —1 8. The coeﬂicIent matrIXA— 0 1 1 O _2 _1 has reduced form 0 O 0 1—3 2 1 O O O 7/6 —5/4 _ O 1 O 0 —20/3 9/2 . _ _ . AR— 0 O 1 0 14/3 _11/2 .W1th 235—01,:35—Bthe general solutlon can be 0 O O 1 —3 2 —7/6 5/4 20/3 —9/2 writtenas X =a _1:/3 +ﬁ 1y; ,so the solution space of AX =0has dimension 1 O 0 1 2=m— rank (A)=6'—4. 4 —3 O 1 1 —3 . . O 2 O 4 —1 —6 10. The c0efﬁc1ent matnx A— 3 _2 O O 4 _1 has reduced form 1; 2 1 —3 4 0 O 1 0 0 0 *W/s “V; 0 1 0 0 to V . . AR : O O 1 O. “AD 4‘: . Wlth x5 = 04,235 = B the general solut10n can be 0 O O 1 47/“, 46/5 '95”: V: “Wk, .18, itte asX “V” + V; th lt' fAX Oh d" ‘ wr n :04 ,so es0u10nsa = 27/“) 9/5 p ce 0 as 1mens10n 1 0 O 1 =m—rank(A)=6—4 2 O 0 O —4 0 1 1 0 2 O O 0 —1 1 —1 12. The coefﬁcient matrixA= O O 1 —4 O O O 1 has reduced form 0 1 —1 1 O O O O 01 O O —1 1 —1 O 1 O O O O —3 7/2 —1/2 0 1 O O O —1/2 1/2 —1/2 AR: 0 0 1 0 O —2/3 2/3 —1 . With 335 =a,:1:7 =B,xg =7the general 0 O O 1 O —1/6 1/6 —1/2 0 n n n1_q/o '2/0 .1/0 (min/meal) 3 —7/2 1/2 1/2 —1/2 1/2 2/3 —2/3 1 solution can be written as X = a + ,3 + 7 , so the solution 1 0 0 0 1 0 0 0 1 space of AX = 0 has dimension 3 = m— rank (A) = 8 — 5. Section 7.6 6. The solution Space of AX = 0 has dimension 2 , as shown by the two arbitrary parameters in the general solution. 8. The solution space of AX = 0 has dimension 2, as shown by the two arbitrary parameters in the general solution. 10. The solution space of AX = 0 has dimension 2, as shown by the two arbitrary parameters in the general solution. 12. The solution space of AX = O has dimension 3, as shown by the three arbitrary parameters in the general solution. Section 7.7 4—231051 2. The augmented matrix is 1 0 0 _3 f 8 . The reduced form of this matrix is 2—301316 1 0 0 —3 E 8 0 1 0 _7/3 3 0 . Since rank (A) = rank (AEB) the system has solutions which 0 0 1 52/9 3 —31/3 8 3 can be expressed as X = 0 + a 7/3 where a is arbitrary —31/3 —52/ 9 ' 0 1 j 2 0 —3 E 1 8. The augmented matrix is 1 _1 1 g 1 . The reduced form of this matrix is 2 —4 1_ 5 2 1 0 0 3/4 0 1 0 _1/12 . Since rank (A) = rank (AB) = number of unknowns = 3, the 0 0 1 1/6 3/4 system has a unique solution X = —1 / 12 1/6 3—23—1 10. The augmented matrix is t 4 3 1 4 ). The reduced form of this matrix is < 1 0 3 5/17 . Since rank (A) = rank (AB) = number of unknowns = 2, the system 0 1 S 16/17 . . _ 5/17 has a unique solution X — < 16/17 12. The augmented matrix is 2 _6 1 g _5 . The reduced form of this matrix is 1 0 0 E ——137/48 0 1 0 f 1/6 . Since rank (A) 2 rank (A53) 2 number of unknowns = 3, the 0 0 1 S 41/24 —l37/48 system has a unique solution X = 1/6 41/24 ——6 2 -1 1 E 0 14. The augmented matrix is 1 4 0 _1 _5 . The reduCed form of this matrix 1 1 1 -—7 E 0 is 1 0 0 21/23 S -—15/23 0 l 0 —ll/23 3 —25/23 . Since rank (A) 2 rank (ASE) the System has solutions 0 0 1 —171/23 3 40/23 -15/23 —21/23 which can be expressed X = ‘25/23 11/23 40/23 + a 171/23 where a is arbitrary. 0 l ...
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AME525_Homework_05_Solutions_101007 - 6 —1 1 O O 10 O O...

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