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quiz x solutions - Math 660 GEOMETRY I SOLUTION FOR QUIZ...

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Math 660 GEOMETRY I SOLUTION FOR QUIZ – X (11/26) December 5 (Thu), 2013 Instructor: Yasuyuki Kachi Line #: 26471. [I] (15pts) Let p 1 : 2 : 0 : 0 : 0 P , p 7 : 0 : 2 : 0 : 0 P , p 2 : 0 : 0 : 1 : 0 P , p 3 : 0 : 0 : 0 : 1 P be four points inside P 4 ( X 0 : X 1 : X 2 : X 3 : X 4 ) . (1a) The answer is ‘False’. None of the three out of these four points are collinear. (1b) The answer is ‘False’. These four points are not co-2-planar. Reason: Form the matrix out of those four coordinate readings: 1 2 0 0 0 7 0 2 0 0 2 0 0 1 0 3 0 0 0 1 . Clearly the rank of this matrix is 4. Indeed, it has a 4 × 4 minor v v v v v v v 2 0 0 0 0 2 0 0 0 0 1 0 0 0 0 1 v v v v v v v which is non-zero. So the criterion (4b) applies, with n = 4 and k = 4, in ‘Fact’ at the end of this solution sheet. (2) The answer is ‘True’. This is the special case of (1), with n = 4 and k = 4, in ‘Fact’ at the end of this solution sheet. 1
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(3) Suppose the defning equation oF the linear span is A 0 X 0 + A 1 X 1 + A 2 X 2 + A 3 X 3 + A 4 X 4 = 0. Substitute each oF p X 0 : X 1 : X 2 : X 3 : X 4 P = p 1 : 2 : 0 : 0 : 0 P , p X 0 : X 1 : X 2 : X 3 : X 4 P = p 7 : 0 : 2 : 0 : 0 P , p X 0 : X 1 : X 2 : X 3 : X 4 P = p 2 : 0 : 0 : 1 : 0 P , and p X 0 : X 1 : X 2 : X 3 : X 4 P = p 3 : 0 : 0 : 0 : 1 P , into the above equation. Then the outcome is A 0 2 A 1 = 0 , 7 A 0 2 A 2 = 0 , 2 A 0 + A 3 = 0 . 3 A 0 A 4 = 0 . You can solve this easily: Substitute A 0 = 1. Then A 1 = 1 2 , A 2 = 7 2 , A 3 = 2, A 4 = 3. Substitute these data back into A 0 X 0 + A 1 X 1 + A 2 X 2 + A 3 X 3 + A 4 X 4 = 0.
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