Instructor's Solutions Manual for Linear Algebra with by Otto and Bretscher and Kyle Burke

By Otto and Bretscher and Kyle Burke

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0 a 0 b , where a, b, and c are arbitrary. 40 We will usually have rref(A) =  0 0  0 1 0 0  0 0 . 41 If Ax = b is a “random” system, then rref(A) will usually be  0 0 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  (by Exercise 39), so that the system will have infinitely many solutions (x4 is a free variable). 0 0 1 c . b] will usually be  . 0    . , so that the system is inconsistent. 0  .

Chapter 1              20 16 12 8 4 0 ducks Thus, we find our solutions for  sparrows  :  0  .  15  ,  30  ,  45  ,  60  and  75 . 73 We let x1 be the number of sheep, x2 be the number of goats, and x3 be the number of hogs. We can then 1 4 7 use the two equations   2 x1 + 3 x2 + 2 x3 = 100 and x1 + x2 + x3 = 100 to generate the following augmented matrix: . 4 7. 1  2 3 2 . 100  . 1 1 1.. 100 then reduce it to  1 0 .. 0 − 13 5 . 18 . 1 5 .  40  . 60  40 + 13 5 s  With this, we see that our solutions will be of the form  60 − 18 5 s .

If we let b = c = 1. Then, a = 2−3(1)+9(1) = 58 . 32 For this problem, we set up the same three equations as in Exercise 30. However, here, we  must enforce that  −2 −2 −2 our matrix, A, contains no zero entries. One possible solution to this problem is the matrix  3 1 2 . 33 The ith component of Ax is [0 0 . . 1 . . 0]   = xi . )  xi    ... xn  Therefore, Ax = x. 34 a Ae1 =  d  , Ae2 =  e , and Ae3 =  f . k h g   1 b Be1 = [v1 v2 v3 ]  0  = 1v1 + 0v2 + 0v3 = v1 . 0 Likewise, Be2 = v2 and Be3 = v3 .

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