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Unformatted text preview: Assignment 3 Math 417 — Winter 2009 Due January 30 § 2.1#6. Yes, it is linear. Its matrix is 1 4 2 5 3 6 . § 2.1#8. The easiest way to find the inverse of a linear transformation is to invert its matrix. So we compute, by rowreduction (saving space by doing two steps together at the second arrow), 1 7 1 0 3 20 0 1 = ⇒ 1 7 1 1 3 1 = ⇒ 1 0 20 7 0 1 3 1 . So the inverse transformation (written with x 1 ,x 2 as the input variables and y 1 ,y 2 as the output) is y 1 = 20 x 1 + 7 x 2 , y 2 = 3 x 1 x 2 . § 2.1#10. Following the method from class, we rowreduce the given matrix augmented by the identity matrix: 1 2 1 0 4 9 0 1 = ⇒ 1 2 1 0 1 4 1 = ⇒ 1 0 9 2 0 1 4 1 . From the last two columns we read off the inverse matrix 9 2 4 1 . § 2.1#12. Row reducing 1 k 1 0 0 1 0 1 takes just one step — subtract k times the bottom row from the top row. So the inverse matrix is 1 k 1 . 1 Remark. Don’t expect inverses to be this nice in general. If you start with a matrix of integers, the inverse may well be full of nasty fractions as part (b) of the following exercise shows. § 2.1#14. The quickest solution here is to use exercise 13 of this section. (a) Using part (a) of problem 13, the matrix 2 3 5 k is invertible exactly when 2 k 15 6 = 0, or k 6 = 15 / 2. (b) We assume that the constant k is an integer, as per the homework instructions. Using part (b) of problem 13, the inverse is given by the formula 2 3 5 k 1 = 1 2 k 15 k 3 5 2 . In order for this to be a matrix of integers, the integer 2 k 15 must divide evenly into the integers k , 3, 5, and 2. The only integers that divide evenly into 3, 5, and 2 are the integers +1 and 1 (which divide evenly into any integer). So, in order for the inverse to be a matrix of integers, we must have 2 k 15 = ± 1. Equivalently, either k = 7 or k = 8. Another approach is to use the method presented in class. We setup the augmented matrix and reduce 2 3 1 0 5 k 0 1 = ⇒ 1 3 / 2 1 / 2 0 5 k 1 = ⇒ 1 3 / 2 1 / 2 k 15 / 2 5 / 2 1 At this point we see that the matrix will be invertible only when k 6 = 15 / 2, which answers part (a). Assuming that k 6 = 15 / 2, write C = 1 / ( k 15 / 2) = 2 / (2 k 15), and keep solving for the inverse 1 3 / 2 1 / 2 k 15 / 2 5 / 2 1 = ⇒ 1 3 / 2 1 / 2 1 5 C/ 2 C = ⇒ 1 0 1 / 2 + 15...
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This note was uploaded on 10/13/2009 for the course MATH 417 taught by Professor Eveet during the Spring '09 term at University of MichiganDearborn.
 Spring '09
 eveet
 Math, Linear Algebra, Algebra

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