Math 308 Section 3.1.pdf - 2017-6-6 UW Common Math 308...

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Unformatted text preview: 2017-6-6 UW Common Math 308 Section 3.1 YIRAN CHEN Math 308, section G, Spring 2017 Instructor: Amos Turchet WebAssign UW Common Math 308 Section 3.1 (Homework) Current Score : 35 / 35 Due : Thursday, April 27 2017 11:00 PM PDT The due date for this assignment is past. Your work can be viewed below, but no changes can be made. Important! Before you view the answer key, decide whether or not you plan to request an extension. Your Instructor may not grant you an extension if you have viewed the answer key. Automatic extensions are not granted if you have viewed the answer key. Request Extension 1. 2/2 points | Previous AnswersHoltLinAlg1 3.1.004. Let T(x) = Ax for the given matrix A, and find T(u1) and T(u2) for the given u1 and u2. −2 A = 4 −2 0 −1 −2 , u1 = 0 −1 −1 0 5 9 , u2 = 7 −4 3 44 T(u1) = -1 -5 [44;­1;­5] [12;­13;­10] 12 T(u2) = -13 -10 Solution or Explanation −2 T(u1) = Au1 = 4 −2 0 −1 −2 0 −1 −1 0 44 9 = −1 , T(u2) = Au2 = 0 −1 −2 7 = −13 −5 0 −1 −1 3 −4 −2 4 −2 5 12 −10 2. 1/1 points | Previous AnswersHoltLinAlg1 3.1.007. Determine if the given vector is in the range of T(x) = Ax, where A = y = 1 −2 0 3 . 2 1 3 7 y is in the range of T. y is not in the range of T. Solution or Explanation We consider T(x) = Ax = y, and row­reduce the corresponding augmented matrix: 1 −2 0 3 3 2 1 7 −3R1 + R2 ~ R2 1 −2 0 0 3 8 1 −2 Since there exists a solution x to Ax = y, y is in the range of T. 1/8 2017-6-6 UW Common Math 308 Section 3.1 3. 1/1 points | Previous AnswersHoltLinAlg1 3.1.010. Suppose that a linear transformation T satisfies 6 T(u1) = 1 −1 , T(u2) = 1 . −5 7 Find T(4u1 − 3u2). T(4u1 − 3u2) = 21 -7 -41 [21;­7;­41] Solution or Explanation 6 T(4u1 − 3u2) = 4T(u1) − 3T(u2) = 4 1 21 −1 − 3 1 = −7 −5 7 −41 4. 5/5 points | Previous AnswersHoltLinAlg1 3.1.013. Determine if the given function is a linear transformation. T(x1, x2) = (7x1 + x2, −2x1 + 9x2) The function is a linear transformation. The function is not a linear transformation. If so, identify the matrix A such that T(x) = Ax. (If the function is not a linear transformation, enter DNE into any cell.) A = 7 1 -2 9 [7,1; ­2,9] If not, explain why not. The function is a linear transformation. The function is not a linear transformation, since there exist numbers a, b, c, and d such that T(a + c, b + d) ≠ T(a, b) + T(c, d). The function is not a linear transformation, since there exist numbers a, b, c, and d such that T(a + c, b + d) = T(a, b) + T(c, d). The function is not a linear transformation, since there exist numbers a, b, and c such that T(a(b, c)) ≠ aT(b, c). The function is not a linear transformation, since there exist numbers a, b, and c such that T(a(b, c)) = aT(b, c). Solution or Explanation Linear transformation, with A = 7 1 −2 9 . 2/8 2017-6-6 UW Common Math 308 Section 3.1 5. 5/5 points | Previous AnswersHoltLinAlg1 3.1.016. Determine if the given function is a linear transformation. T(x1, x2, x3) = (−2x2, 7x3) The function is a linear transformation. The function is not a linear transformation. If so, identify the matrix A such that T(x) = Ax. (If the function is not a linear transformation, enter DNE into any cell.) A = 0 0 -2 0 0 7 [0,­2,0; 0, 0, 7] If not, explain why not. The function is a linear transformation. The function is not a linear transformation, since there exist numbers a, b, c, d, e, and f such that T(a + d, b + e, c + f) ≠ T(a, b, c) + T(d, e, f). The function is not a linear transformation, since there exist numbers a, b, c, d, e, and f such that T(a + d, b + e, c + f) = T(a, b, c) + T(d, e, f). The function is not a linear transformation, since there exist numbers a, b, c, and d such that T(a(b, c, d)) ≠ aT(b, c, d). The function is not a linear transformation, since there exist numbers a, b, c, and d such that T(a(b, c, d)) = aT(b, c, d). Solution or Explanation Linear transformation, with A = 0 −2 0 0 0 7 . 6. 2/2 points | Previous AnswersHoltLinAlg1 3.1.022. Let T(x) = Ax for the given matrix A. Determine if T is one­to­one and if T is onto. A = 4 2 12 6 T is one­to­one. T is onto. T is both one­to­one and onto. T is neither one­to­one nor onto. Solution or Explanation We consider T(x) = Ax = b, and row­reduce the corresponding augmented matrix: 4 2 b1 12 6 b2 −3R1 + R2 ~ R2 4 2 b1 0 0 −3b1 + b2 If −3b1 + b2 ≠ 0, there does not exist a unique solution x to Ax = b. By The Big Theorem ­ Version 2, T is neither one­to­one nor onto. 3/8 2017-6-6 UW Common Math 308 Section 3.1 7. 2/2 points | Previous AnswersHoltLinAlg1 3.1.025. Let T(x) = Ax for the given matrix A. Determine if T is one­to­one and if T is onto. A = 1 −5 −4 19 2 −14 T is one­to­one. T is onto. T is both one­to­one and onto. T is neither one­to­one nor onto. Solution or Explanation Since n = 3 > m = 2, by Theorem 3.7 T is not onto. To determine if T is one­to­one, we row­reduce the corresponding augmented matrix: 1 −5 0 4R1 + R2 19 0 −2R1 + R3 ~ 2 −14 0 −4 R2 R3 1 −5 0 0 −1 0 −4R2 + R3 0 −4 0 ~ R3 1 −5 0 0 −1 0 0 0 0 Since T(x) = Ax = 0 has only the trivial solution, by Theorem 3.5 T is one­to­one. 8. 2/2 points | Previous AnswersHoltLinAlg1 3.1.028. Let T(x) = Ax for the given matrix A. Determine if T is one­to­one and if T is onto. A = 1 2 −4 3 7 −7 −2 −4 3 T is one­to­one. T is onto. T is both one­to­one and onto. T is neither one­to­one nor onto. Solution or Explanation We consider T(x) = Ax = b, and row­reduce the corresponding augmented matrix: 1 2 −4 b1 −3R1 + R2 7 −7 b2 2R1 + R3 ~ −2 −4 3 b3 3 R2 R3 b1 1 2 −4 0 1 5 −3b1 + b2 0 0 −5 2b1 + b3 Since there exists a unique solution x to Ax = b. By The Big Theorem ­ Version 2, T is both one­to­one and onto. 4/8 2017-6-6 UW Common Math 308 Section 3.1 9. 3/3 points | Previous AnswersHoltLinAlg1 3.1.031. Suppose that T(x) = Ax for the given A. Sketch a graph of the image under T of the unit square in the first quadrant of R2. A = 1 −6 8 1 Solution or Explanation 5/8 2017-6-6 UW Common Math 308 Section 3.1 10.3/3 points | Previous AnswersHoltLinAlg1 3.1.034. Find an example that meets the given specifications. A linear transformation T : R2 → R3 such that T T(x) = 0 1 0 5 0 6 0 1 1 = 5 . 6 x [0,1;0,5;0,6] Solution or Explanation 0 1 T(x) = 0 5 x 0 6 11.3/3 points | Previous AnswersHoltLinAlg1 3.1.037. Find an example that meets the given specifications. A linear transformation T : R2 → R2 such that T 1 3 1 = 0 13 and T 1 4 = −11 8 . -3 T(x) = 4 1 x [1,­3;4,1] Solution or Explanation T(x) = 1 −3 4 1 x 6/8 2017-6-6 UW Common Math 308 Section 3.1 12.1/1 points | Previous AnswersHoltLinAlg1 3.1.040. Determine if the statement is true or false, and justify your answer. The range of a linear transformation must be a subset of the domain. True, by definition of the domain. True, by definition of the range. x1 False. For instance T : R2 → R2 defined by T = x1 + x2 has range(T) = R, which is not a subset of R2, the domain of x2 T. False. For instance T : R2 → R defined by T False. For instance T : R2 → R defined by T x1 = x1 + x2 has range(T) = R, which is not a subset of R2, the domain of T. x2 x1 = x1 + x2 has range(T) = R2, which is not a subset of R, the domain of T. x2 Solution or Explanation False. For instance T : R2 → R defined by T x1 x2 = x1 + x2 has range(T) = R, which is not a subset of R2, the domain of T. 13.1/1 points | Previous AnswersHoltLinAlg1 3.1.046. Determine if the statement is true or false, and justify your answer. If T1(x) and T2(x) are onto linear transformations from Rn to Rm, then so is W(x) = T1(x) + T2(x). True, by the definition of linear transformation. True, by the definition of linear transformation and the definition of onto. False. Consider T2(x) = T1(x), where T1 is onto. False. Consider T2(x) = −T1(x), where T1 is one­to­one. False. Consider T2(x) = −T1(x), where T1 is onto. Solution or Explanation False. W will be linear, but not necessarily onto. Consider T2(x) = −T1(x), where T1 is onto. 7/8 2017-6-6 UW Common Math 308 Section 3.1 14.4/4 points | Previous AnswersHoltLinAlg1 3.1.049. A linear transformation T : R2 → R2 is called a dilation if T(x) = rx for r > 1. (It is called a contraction if 0 < r < 1.) (a) Find the matrix A such that T(x) = Ax. A = r 0 0 r [r, 0; 0, r] (b) Let r = 2, and then sketch the graphs of x = 3 −1 and T(x). Solution or Explanation (a) A = r 0 0 r (b) 8/8 ...
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