# 36 by exercise 35 the i th column of a is ae i for i

• Notes
• 51
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 41 - 44 out of 51 pages.

36. By Exercise 35, the i th column of A is Ae i , for i = 1 , 2 , 3. Therefore, A = 1 4 7 2 5 8 3 6 9 . 37. We have to solve the system x 1 + 2 x 2 = 2 x 3 = 1 or x 1 = 2 - 2 x 2 x 3 = 1 . Let x 2 = t . Then the solutions are of the form x 1 x 2 x 3 = 2 - 2 t t 1 , where t is an arbitrary real number. 38. We will illustrate our reasoning with an example. We generate the “random” 3 × 3 matrix A = 0 . 141 0 . 592 0 . 653 0 . 589 0 . 793 0 . 238 0 . 462 0 . 643 0 . 383 . Since the entries of this matrix are chosen from a large pool of numbers (in our case 1000, from 0.000 to 0.999), it is unlikely that any of the entries will be zero (and even less likely that the whole first column will consist of zeros). This means that we will usually be able to apply Steps 2 and 3 of the Gauss-Jordan algorithm to turn the first 41
Chapter 1 ISM: Linear Algebra column into 1 0 0 ; this is indeed possible in our example: 0 . 141 0 . 592 0 . 653 0 . 589 0 . 793 0 . 238 0 . 462 0 . 643 0 . 383 -→ 1 4 . 199 4 . 631 0 - 1 . 680 - 2 . 490 0 - 1 . 297 - 1 . 757 . Again, it is unlikely that any entries in the second column of the new matrix will be zero. Therefore, we can turn the second column into 0 1 0 . Likewise, we will be able to clear up the third column, so that rref( A ) = 1 0 0 0 1 0 0 0 1 . We summarize: As we apply Gauss-Jordan elimination to a random matrix A (of any size), it is unlikely that we will ever encounter a zero on the diagonal. Therefore, rref( A ) is likely to have all ones along the diagonal. 39. We will usually get rref( A ) = 1 0 0 a 0 1 0 b 0 0 1 c , where a, b , and c are arbitrary. 40. We will usually have rref( A ) = 1 0 0 0 1 0 0 0 1 0 0 0 . (Compare with the summary to Exercise 38.) 41. If Ax = b is a “random” system, then rref( A ) will usually be 1 0 0 0 1 0 0 0 1 , so that we will have a unique solution. 42. If Ax = b is a “random” system of three equations with four unknowns, then rref( A ) will usually be 1 0 0 a 0 1 0 b 0 0 1 c (by Exercise 39), so that the system will have infinitely many solutions ( x 4 is a free variable). 42
ISM: Linear Algebra Section 1.3 43. If Ax = b is a “random” system of equations with three unknowns, then rref[ A . . . b ] will usually be 1 0 0 . . .0 0 1 0 . . .0 0 0 1 . . .0 0 0 0 . . .1 , so that the system is inconsistent. 44. Let E = rref( A ) , and note that all the entries in the last row of E must be zero, by the definition of rref. If c is any vector in R n whose last component isn’t zero, then the system Ex = c will be inconsistent. Now consider the elementary row operations that transform A into E, and apply the opposite operations, in reversed order, to the augmented matrix E . . . c . You end up with an augmented matrix A . . . b that represents an inconsistent system Ax = b, as required. 45. Write A = [ v 1 v 2 . . . v m ] and x = x 1 . . . x m . Then A ( kx ) = [ v 1 . . . v m ] kx 1 . . . kx m = kx 1 v 1 + · · · + kx m v m and k ( Ax ) = k ( x 1 v 1 + · · · + x m v m ) = kx 1 v 1 + · · · + kx m v m . The two results agree, as claimed.
• • • 