Sect3 2

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Sect3 2

  1. 1. SECTION 3.2 MATRICES AND GAUSSIAN ELIMINATION Because the linear systems in Problems 1-10 are already in echelon form, we need only start at the end of the list of unknowns and work backwards. 1. Starting with x3 = 2 from the third equation, the second equation gives x2 = 0, and then the first equation gives x1 = 1. 2. Starting with x3 = −3 from the third equation, the second equation gives x2 = 1, and then the first equation gives x1 = 5. 3. If we set x3 = t then the second equation gives x2 = 2 + 5t , and next the first equation gives x1 = 13 + 11t. 4. If we set x3 = t then the second equation gives x2 = 5 + 7t , and next the first equation gives x1 = 35 + 33t. 5. If we set x4 = t then the third equation gives x3 = 5 + 3t , next the second equation gives x2 = 6 + t , and finally the first equation gives x1 = 13 + 4t. 6. If we set x3 = t and x4 = −4 from the third equation, then the second equation gives x2 = 11 + 3t , and next the first equation gives x1 = 17 + t. 7. If we set x3 = s and x4 = t , then the second equation gives x2 = 7 + 2 s − 7t , and next the first equation gives x1 = 3 − 8s + 19t. 8. If we set x2 = s and x4 = t , then the second equation gives x3 = 10 − 3t , and next the first equation gives x1 = −25 + 10 s + 22t. 9. Starting with x4 = 6 from the fourth equation, the third equation gives x3 = −5, next the second equation gives x2 = 3, and finally the first equation gives x1 = 1. 10. If we set x3 = s and x5 = t , then the third equation gives x4 = 5t , next the second equation gives x2 = 13s − 8t , and finally the first equation gives x1 = 63s − 16t. In each of Problems 11-22, we give just the first two or three steps in the reduction. Then we display a resulting echelon form E of the augmented coefficient matrix A of the given linear system, and finally list the resulting solution (if any). The student should understand that the
  2. 2. echelon matrix E is not unique, so a different sequence of elementary row operations may produce a different echelon matrix. 11. Begin by interchanging rows 1 and 2 of A. Then subtract twice row 1 both from row 2 and from row 3. 1 3 2 5  E = 0 1 0 −2  ;   x1 = 3, x2 = −2, x3 = 4 0 0 1 4    12. Begin by subtracting row 2 of A from row 1. Then subtract twice row 1 both from row 2 and from row 3. 1 −6 −4 15  E = 0 1 0 −3 ;   x1 = 5, x2 = −3, x3 = 2 0 0 1 2    13. Begin by subtracting twice row 1 of A both from row 2 and from row 3. Then add row 2 to row 3. 1 3 3 13 E = 0 1 2 3  ;   x1 = 4 + 3t , x2 = 3 − 2t , x3 = t 0 0 0 0    14. Begin by interchanging rows 1 and 3 of A. Then subtract twice row 1 from row 2, and three times row 1 from row 3.  1 −2 − 2 −9  E = 0 0 1 7  ;   x1 = 5 + 2t , x2 = t , x3 = 7 0 0 0 0    15. Begin by interchanging rows 1 and 2 of A. Then subtract three times row 1 from row 2, and five times row 1 from row 3. 1 1 1 1  E =  0 1 3 3 .   The system has no solution. 0 0 0 1  
  3. 3. 16. Begin by subtracting row 1 from row 2 of A. Then interchange rows 1 and 2. Next subtract twice row 1 from row 2, and five times row 1 from row 3.  1 −4 − 7 6  E = 0 1  2 0 .  The system has no solution. 0 0 0 1   17. x1 − 4 x2 − 3 x3 − 3 x4 = 4 2 x1 − 6 x2 − 5 x3 − 5 x4 = 5 3x1 − x2 − 4 x3 − 5 x4 = − 7 [The first printing of the textbook had a misprinted 2 (instead of 4) as the right-hand side constant in the first equation.] Begin by subtracting twice row 1 from row 2 of A, and three times row 1 from row 3.  1 −4 − 3 −3 4  E = 0 1 0 −1 −4  ;   x1 = 3 − 2t , x2 = −4 + t , x3 = 5 − 3t , x4 = t 0 0 1 3 5    18. Begin by subtracting row 3 from row 1 of A. Then subtract 3 times row 1 from row 2, and twice row 1 from row 3. 1 −2 −4 −13 −8  E = 0 0 1  4 3 ;  x1 = 4 + 2 s − 3t , x2 = s, x3 = 3 − 4t , x4 = t 0 0 0  0 0  19. Begin by interchanging rows 1 and 2 of A. Then subtract three times row 1 from row 2, and four times row 1 from row 3.  1 −2 5 −5 −7  E =  0 1 −2 3 5  ;   x1 = 3 − s − t , x2 = 5 + 2s − 3t , x3 = s, x4 = t 0 0 0 0 0    20. Begin by interchanging rows 1 and 2 of A. Then subtract twice row 1 from row 2, and five times row 1 from row 3.  1 3 2 −7 3 9  E = 0 1 3 −7 2 7  ;   x1 = 2 + 3t , x2 = 1 + s − 2t , x3 = 2 + 2s , x4 = s, x5 = t  0 0 1 −2 0 2   
  4. 4. 21. Begin by subtracting twice row 1 from row 2, three times row 1 from row 3, and four times row 1 from row 4. 1 1 1 0 6 0 1 5 1 20  E =  ; x1 = 2, x2 = 1, x3 = 3, x4 = 4 0 0 1 0 3   0 0 0 1 4 22. Begin by subtracting row 4 from row 1. Then subtracting twice row 1 from row 2, four times row 1 from row 3, and three times row 1 from row 4.  1 −2 − 4 0 −9  0 1 6 1 21 E =  ; x1 = 3, x2 = −2, x3 = 4, x4 = −1 0 0 1 0 4   0 0 0 1 −1 23. If we subtract twice the first row from the second row, we obtain the echelon form 3 2 1  E =   0 0 k − 2  of the augmented coefficient matrix. It follows that the given system has no solutions unless k = 2, in which case it has infinitely many solutions given by x1 = 1 (1 − 2t ), x2 = t. 3 24. If we subtract twice the first row from the second row, we obtain the echelon form 3 2 0 E =   0 k − 4 0  of the augmented coefficient matrix. It follows that the given system has only the trivial solution x1 = x2 = 0 unless k = 4, in which case it has infinitely many solutions given by x1 = − 2 t , x2 = t. 3 25. If we subtract twice the first row from the second row, we obtain the echelon form 3 2 11  E =   0 k − 4 −1 of the augmented coefficient matrix. It follows that the given system has a unique solution if k ≠ 4, but no solution if k = 4.
  5. 5. 26. If we first subtract twice the first row from the second row, then interchange the two rows, and finally subtract 3 times the first row from the second row, then we obtain the echelon form 1 1 k − 2  E =   0 1 3k − 7  of the augmented coefficient matrix. It follows that the given system has a unique solution whatever the value of k. 27. If we first subtract twice the first row from the second row, then subtract 4 times the first row from the third row, and finally subtract the second row from the third row , we obtain the echelon form 1 2 1 3  E =  0 5 5 1   0 0 0 k − 11   of the augmented coefficient matrix. It follows that the given system has no solution unless k = 11, in which case it has infinitely many solutions with x3 arbitrary. 28. If we first interchange rows 1 and 2, then subtract twice the first row from the second row, next subtract 7 times the first row from the third row, and finally subtract twice the second row from the third row , we obtain the echelon form 1 2 1 b  0 −5 1 E =  a − 2b   0 0 0 c − 2a − 3b    of the augmented coefficient matrix. It follows that the given system has no solution unless c = 2a + 3b, in which case it has infinitely many solutions with x3 arbitrary. 29. In each of parts (a)-(c), we start with a typical 2 × 2 matrix A and carry out two row successive operations as indicated, observing that we wind up with the original matrix A.  s t  c R2  s t  (1/ c ) R 2 s t  (a) A =   →  cu cv  → u v  = A u v      s t  SWAP ( R1, R 2) u v  SWAP ( R1, R 2) s t  (b) A =   → s t  → u v  = A u v      s t  c R1+ R 2  u v  ( − c ) R1+ R 2 s t  (c) A =   → cu + s cv + t  → u v  = A u v     
  6. 6. Since we therefore can "reverse" any single elementary row operation, it follows that we can reverse any finite sequence of such operations — on at a time — so part (d) follows. 30. (a) This part is essentially obvious, because a multiple of an equation that is satisfied is also satisfied, and the sum of two equations that are satisfied is one that is also satisfied. (b) Let us write A1 = B1 , B 2 , , B n , B n+1 = A 2 where each matrix B k +1 is obtained from B k by a single elementary row operation (for k = 1, 2, , n ). Then it follows by n applications of part (a) that every solution of the system LS1 associated with the matrix A1 is also a solution of the system LS2 associated with the matrix A2. But part (d) of Problem 29 implies that A1 also can be obtained by A2 by elementary row operations, so by the same token every solution of LS2 is also a solution of LS1.

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