m307_mid1_05_ans[1] - UBC Mathematics 307 Section 921...

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Unformatted text preview: UBC Mathematics 307 Section 921 Midterm Exam I (Jun 9, 2005) Instructor: Yue-Xian Li Answers I. (10 marks) Find a system Ax = b (where A is 2 3 and b is 2 1) whose general solution is . 32. 12 x = . ..2 . + s . ..3 . . 01 2. . u . 10 ..21 Soln: . v . = for any . = 0. 6 0 . 3. ..2. w Or 2. . u . 111 1 . v . = ..6 for any . = 0. 0 . 3. 2. w II. (20 marks) 2. . 32. u 1231 . . v . 7 Let A = . 2453 5 , b = . 2 5 , and x = . . . w . 1222 ..5 z (1) Find the lower triangular matrix L with 1's on the diagonal and the echelon form U such that A = LU. (2) Describe the relationship between the nullspace of A (denoted by N (A)) and the general solution of Ax = 0. Express N (A) in parametric vector form. (3) Find the value of . that makes the system Ax = b consistent. (4) Write the general solution of Ax = b for the vector b found in (3) that makes it consistent. Soln: 23. . 1 1231 (1) A = LU = . 21 5. 00 ..11 . . 111 0000 (2) The general solution of Ax = 0 constitutes the nullspace of A. It contains all possible linear relations of the columns of A. These relations are: v2 =2v1 and v4 = ..v3 +4v1 . 23. . 24 67. 7 ..10 67. 7 Thus, N (A)= s + t. 45. 5 0 ..1 0 ..1 (3) b. = ..5 - . - (2 - 2 )= . - 7 = 0. Thus, . = 7 makes the system consistent. 3 . 3232. ..292 4 . 0 7. ..1 7. 0 . . 76767 (4) The general solution is x = xp + xh = xp + N (A)= + s + t , . 54545 12 0 ..1 00 ..1 2. ..29 67 0 67 where xp = is found by solving the following system . 12 . 0 . . . . u . . . . . 1 0 2 0 3 ..1 1 1 . . . 0 . . = . b1 b2 - 2b1 . = . 7 ..12 . . 0 0 0 0 . w . 0 0 0 III. (20 marks) True or false. Show reason if true. Show reason or counter example if false. (1) If A is invertible with inverse A..1 , so is AT with inverse (A..1 )T . True. Because AT (A..1 )T =(A..1 A)T = IT = I and (A..1 )T AT =(AA..1 )T = IT = I. (2) If Ax = Ay, then x = y. False. Because if N (A) 6 y 2N (A), Ax = Ay = 0 but x = y. = f0g, then if x, 6 . . . . . 1 ..112 12 For example, if A =, A = A = 0, but 6=. 0012 12 (3) The set of all vectors in R3 whose dot product with a given vector v is zero forms a subspace of R3 . It de nes a plane that is orthogonal to the vector v. (This set can be expressed as S = fx . R3 : such that x v = 0 for a given v . R3 g.) True. Proof: If a, b . S, then av = 0 and bv = 0. Since ( a+ b)v = (av)+ (bv) =0, thus a + b . S. (4) The second row of EA is equal to the last row of A subtract the rst row of A. Here, 2. ..1 ..1 ..1 A is a 3 n matrix, while E = . 101 5 . 111 . 323. . ..1 ..1 ..1 r1 ..r1 - r2 - r3 False. Since EA = . 101 5. r2 . = . r1+ r3 5. 111 r3 r1+ r2+ r3 (5) The last column of AE is equal to the sum of all the columns of A. Here, A is a m 3 . ..1 0 1 . matrix, while E = . 0 1 1 ..1 1 1 5. . ..1 0 1 . True. Since AE = [c1 c2 c3] . 0 1 1 ..1 1 1 . = [..c1 + c3, c2 - c3, c1 + c2 + c3]. ...
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