Introduction to Linear algebra-Strang-Solutions-Manual_ver13

Their averages are actually innite with row exchanges

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Unformatted text preview: ns are and ,AD4 and U D 4 . 17 26 0 0 1 15 0 0 1 15 0101 0 000 Elimination leaves the diagonal matrix diag.3; 2; 1/ in 3x D 3; 2y D 2; z D 4. Then x D 1; y D 1; z D 4. A.2; W/ D A.2; W/ 3  A.1; W/ subtracts 3 times row 1 from row 2. The average pivots for rand(3) without row exchanges were 1 ; 5; 10 in one experiment— 2 but pivots 2 and 3 can be arbitrarily large. Their averages are actually infinite ! With row exchanges in MATLAB’s lu code, the averages :75 and :50 and :365 are much more stable (and should be predictable, also for randn with normal instead of uniform probability distribution). If A.5; 5/ is 7 not 11, then the last pivot will be 0 not 4. Row j of U is a combination of rows 1; : : : ; j of A. If Ax D 0 then U x D 0 (not true if b replaces 0). U is the diagonal of A when A is lower triangular. The question deals with 100 equations Ax D 0 when A is singular. (a) Some linear combination of the 100 rows is the row of 100 zeros. (b) Some linear combination of the 100 columns is the column of zeros. (c) A very singular matrix has all ones: A D eye(100). A better example has 99 random rows (or the numbers 1i ; : : : ; 100i in those rows). The 100th row could be the sum of the first 99 rows (or any other combination of those rows with no zeros). (d) The row picture has 100 planes meeting along a common line through 0. The column picture has 100 vectors all in the same 99-dimensional hyperplane. Problem Set 2.3, page 63 " # " # " #" #" # 100 100 100 010 010 5 1 0 ; E32 D 0 1 0 ; P D 0 0 1 100 D 001. 1 E21 D 001 071 010 001 100 2 E32 E21 b D .1; 5; 35/ but E21 E32 b D .1; 5; 0/. When E32 comes first, row 3 feels no effect from row 1. " #" #" # " # 100 100 1 0 0 1 0 0 410;010;0 1 0 M D E32 E31 E21 D 4 1 0: 3 001 201 0 2 1 10 2 1 Solutions to Exercises 12 23 23 2 1 1 6 7 E21 6 7 E31 6 4 Elimination on column 4: b D 4 0 5 ! 4 4 5 ! 4 0 0 original Ax D b has become U x D c D .1; 4 ; 10/. Then 1 z D 5; y D 2 ; x D 1 : This solves Ax D .1; 0; 0/. 2 1 3 2 1 3 7 E32 6 7 4 5 ! 4 4 5. The 2 10 back substitution gives 5 Changing a33 from 7 to 11 will change the third pivot from 5 to 9. Changing a33 from 7 to 2 will change the pivot from 5 to no pivot. 2 32 3 23 237 1 4 6 76 7 67 6 Example: 4 2 3 7 5 4 3 5 D 4 4 5. If all columns are multiples of column 1, 237 there is no second pivot. 1 4 7 To reverse E31 , add 7 times row 1 to row 3. The inverse of the elimination matrix 3 2 3 2 8 9 10 11 12 100 100 7 6 7 6 1 E D 4 0 1 0 5 is E D 4 0 1 0 5. 701 701     ab a b MD and M * D . det M * D a.d `b/ b .c `a/ cd c `a d `b reduces to ad b c ! " # 100 M D 0 0 1 . After the exchange, we need E31 (not E21 ) to act on the new row 3. 110 " #" # " # 101 101 201 E13 D 0 1 0 I 0 1 0 I E31 E13 D 0 1 0 : Test on the identity matrix! 001 101 101 # " 122 An example with two negative pivots is A D 1 1 2 . The diagonal entries can 121 change sign during elimination. # " # " rows and 1 2 3 987 also columns T...
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