# 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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**Example text**

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 .