There are no solutions 14 the system reduces to x 2 y

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There are no solutions. 14. The system reduces to x + 2 y = - 2 z = 2 -→ x = - 2 - 2 y z = 2 . Let y = t . x y z = - 2 - 2 t t 2 15. The system reduces to x = 4 y = 2 z = 1 . 16. The system reduces to x 1 + 2 x 2 + 3 x 3 +5 x 5 = 6 x 4 +2 x 5 = 7 -→ x 1 = 6 - 2 x 2 - 3 x 3 - 5 x 5 x 4 = 7 - 2 x 5 . Let x 2 = r, x 3 = s , and x 5 = t . 17
Chapter 1 ISM: Linear Algebra x 1 x 2 x 3 x 4 x 5 = 6 - 2 r - 3 s - 5 t r s 7 - 2 t t 17. The system reduces to x 1 = - 8221 4340 x 2 = 8591 8680 x 3 = 4695 434 x 4 = - 459 434 x 5 = 699 434 . 18. a. No, since the third column contains two leading ones. b. Yes c. No, since the third row contains a leading one, but the second row does not. d. Yes 19. 0 0 0 0 and 1 0 0 0 20. Four, namely 0 0 0 0 , 1 k 0 0 , 0 1 0 0 , 1 0 0 1 ( k is an arbitrary constant.) 21. Four, namely 0 0 0 0 0 0 , 1 k 0 0 0 0 , 0 1 0 0 0 0 , 1 0 0 1 0 0 ( k is an arbitrary constant.) 22. Seven, namely 0 0 0 0 0 0 , 1 a b 0 0 0 , 0 1 c 0 0 0 , 0 0 1 0 0 0 , 1 0 d 0 1 e , 1 f 0 0 0 1 , 0 1 0 0 0 1 . Here, a, b, . . . , f are arbitrary constants. 23. We need to show that the matrix has the three properties listed on page 16. 18
ISM: Linear Algebra Section 1.2 Property a holds by Step 2 of the Gauss-Jordan algorithm (page 17). Property b holds by Step 3 of the Gauss-Jordan algorithm. Property c holds by Steps 1 and 4 of the algorithm. 24. Yes; each elementary row operation is reversible, that is, it can be “undone.” For example, the operation of row swapping can be undone by swapping the same rows again. The operation of dividing a row by a scalar can be reversed by multiplying the same row by the same scalar. 25. Yes; if A is transformed into B by a sequence of elementary row operations, then we can recover A from B by applying the inverse operations in the reversed order (compare with Exercise 24). 26. Yes, by Exercise 25, since rref( A ) is obtained from A by a sequence of elementary row operations. 27. No; whatever elementary row operations you apply to 1 2 3 4 5 6 7 8 9 , you cannot make the last column equal to zero. 28. Suppose ( c 1 , c 2 , . . . , c n ) is a solution of the system a 11 x 1 + a 12 x 2 + · · · + a 1 n x n = b 1 a 21 x 1 + a 22 x 2 + · · · + a 2 n x n = b 2 . . . . . . . . . . To keep the notation simple, suppose we add k times the first equation to the second; then the second equation of the new system will be ( a 21 + ka 11 ) x 1 + · · · +( a 2 n + ka 1 n ) x n = b 2 + kb 1 . We have to verify that ( c 1 , c 2 , . . . , c n ) is a solution of this new equation. Indeed, ( a 21 + ka 11 ) c 1 + · · · + ( a 2 n + ka 1 n ) c n = a 21 c 1 + · · · + a 2 n c n + k ( a 11 c 1 + · · · + a 1 n c n ) = b 2 + kb 1 . We have shown that any solution of the “old” system is also a solution of the “new.” To see that, conversely, any solution of the new system is also a solution of the old system, note that elementary row operations are reversible (compare with Exercise 24); we can obtain the old system by subtracting k times the first equation from the second equation of the new system.

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