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Unformatted text preview: Solutions to Assignment 1 Math 217, Fall 2002 1.1.25 Find an equation involving g , h , and k that makes this augmented matrix correspond to a consistent system: 1 4 7 g 3 5 h 2 5 9 k . To see if this matrix is consistent, we put it in row reduced form. First label the rows R 1 , R 2 , and R 3 . Then 1 4 7 g 3 5 h 2 5 9 k l R 3 → R 3 + 2 R 1 1 4 7 g 3 5 h 3 5 k + 2 g l R 3 → R 3 + R 2 1 4 7 g 3 5 h k + 2 g + h . We see that this matrix will be consistent if and only if k + 2 g + h = 0. 1.1.27 Suppose that the system below is consistent for all possible values of f and g . What can you say about the coefficients c and d ? Justify your answer. x 1 + 3 x 2 = f cx 1 + dx 2 = g. This system corresponds to the augmented matrix: 1 3 f c d g 1 which has row echelon form: 1 3 f d 3 c g cf . If d = 3 c , then this matrix is inconsistent whenever g cf 6 = 0 (take g = 1, f = 0, for instance). Because this matrix is supposed to be consistent for all f and g , we can conclude that d 6 = 3 c . 1.2.20 Choose h and k such that the system below has (a) no solution, (b) a unique solution, and (c) many solutions. x 1 + 3 x 2 = 2 3 x 1 + hx 2 = k. This system corresponds to the augmented matrix: 1 3 2 3 h k . which has row echelon form: A = 1 3 2 h 9 k 6 . (a) If h = 9 and k = 7, then A becomes 1 3 2 0 0 1 , which is inconsistent. Thus for h = 9 and k = 7, the system above has no solution. More generally, this system has a no solution whenever h = 9 and k 6 = 6. (b) If k = 10, then A becomes 1 3 2 0 1 k 6 , and because every column contains a pivot row, this gives a unique solution for all choices of k (for instance, k = 0 works just fine). Formally, the system given above has a unique solution for h = 10, k = 0. More generally, this system has a unique solution for all k and h such that h 6 = 9. (c) Finally, if h = 9 and k = 6, then A becomes 1 3 2 0 0 0 , which is consistent and contains a free column. Thus the system given above has a free variable, and hence, infinitely many solutions. 2 1.2.28 What would you have to know about the pivot columns in an augmented matrix in order to know that the linear system is consistent and has a unique solution? To know that a system has a unique solution, you need to know that it is consistent (so no pivot in the last column of the augmented matrix), and that there are no free variables (so a pivot position in each column of the coefficient matrix). Note that this also means there can not be less rows then columns (see theorem 8, pg. 69)....
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This note was uploaded on 02/22/2012 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.
 Spring '08
 Neely

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