answers1 - Answers to Selected Questions Calculus,...

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Answers to Selected Questions Calculus, Techniques of Optimization with Microeconomics Applications John Hoag Section I 3. a. is one to one b. is not one to one. Consider y = 1. We could have either x = 1 or x = -1. c. is one to one d. is not one to one. For y = 16, any x will work. 4. The function f(n) = 10 n is one to one on the integers. Section III 1. column j = nj j j a a a . . . 2 1 row j = jn j j a a a . . . 2 1 3.b 4 . 1 4 8 . 1 24 6 . 1 2 18 22 2 4 . 8 . 6 4.b. 10 1 7 3 10 4 9 1 0 e. [] 32 , 19 , 12 6.a. 2 + 4 + 8 + 16 = 30 e. a 11 b 13 + a 12 b 23 + a 13 b 33 + a 14 b 43
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7.a. The multiplication is not defined. B is 2 x 1 and A is 2 x 2. c. 19 g. 28 0 7 4 0 1 8 0 2 9.c.1 5x 1 2 + 6x 1 x 2 – 2x 2 2 + 8x 1 x 3 + 3x 3 2 12.a. 5 2 4 4 3 2 3 1 1 13.a. 1 c. π e+ 22 14.a. 3 d. a 11 a 22 a 33 h. 33 32 31 23 22 21 13 12 11 a a a a a a a a a λλ λ = λ 3 |A| 15. n = 7, i = 4, we cross out 3 rows and three columns, the matrix whose determinant we would then find would be 4 x 4. 18. First order minors: each element is a first order minor. -2, 0, 1, 3, -5, 2, -1, -3, 4 Second order minors: 4 3 2 5 = -14 4 1 2 3 = 14 3 1 5 3 = -14 4 3 1 0 = 3 4 1 1 2 = -7 3 1 0 2 = 6
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2 5 1 0 = 5 2 3 1 2 = -7 5 3 0 2 = 10 Third order minor is the determinant of the 3 x 3 matrix, 14. 21. There are 4 1 st order principal minors in a 4 x 4 matrix. There are six 2 nd order principal minors of a 4 x 4, and 4 3 rd order principal minors. 24. If the matrix is symmetric, we the minors of a ij and a ji will be the same. How can we tell? The minor of element a ij requires crossing our row i and column j. The minor of element a ji requires crossing out row j and column i. Because A is symmetric, the transpose of the minor of a ij will be the minor of element a ji . Because |A| = |A’| the two determinants must be the same. If A is not symmetric, then we have nothing to say about how the minors of a ij and a ji are related. 25. Second order cofactors: (-1) 1+1 4 3 2 5 = -14 (-1) 1+2 4 1 2 3 = -14 (-1) 1+3 3 1 5 3 = -14 (-1) 2+1 4 3 1 0 = -3 (-1) 2+2 4 1 1 2 = -7 (-1) 2+3 3 1 0 2 = -6 (-1) 3+1 2 5 1 0 = 5 (-1) 3+2 2 3 1 2 = 7 (-1) 3+3 5 3 0 2 = 10 27.a. |A| = 1 b. |A| = 24 d. |A| = 8 e. –17, -17 (because the second matrix is obtained from the first one by second row and second column so that each time the determinant changes its sign with the same value). g. |A| = 948 j. –17, -25 (because the second matrix is obtained from the first one by making it symmetric as we did for quadratic forms).
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28. Call the new matrix B. To find |B|, use the LaPlace expansion based on the row that has been multiplied by k. Note that to find the minors needed, we cross out the only row that has changed, the row multiplied by k. Thus the cofactors, C ij , will be the same as in the original matrix. Now the determinant is |B| = = k = k|A|.
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answers1 - Answers to Selected Questions Calculus,...

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