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Unformatted text preview: 0 : −1 , into the above equation. Then the outcome is A 0 − 2A 1
7 A0 = 0,
− 2A 2 2 A0 = 0,
+ A3 3 A0 = 0.
− A 4 = 0. You can solve this easily: Substitute A0 = 1. Then
A1 = 1
,
2 A2 = 7
,
2 A3 = −2, A4 = 3. Substitute these data back into
A0 X0 + A1 X1 + A2 X2 + A3 X3 + A4 X4 = 0.
So the ﬁnal answer is
X0 + 1
X1 +
2 7
X2 − 2 X3 + 3 X4 = 0.
2 Or, equivalently,
2 X0 + X1 + 7 X2 − 4 X3 + 6 X4 = 0.
2 and [II] (15pts) Let 1:0:0:2:6, 2 : 0 : −1 : 0 : −2 , 1:2:0:0:2 be three points inside P4 (X0 X1 :X2 :X3 :X4 ) .
(1) The answer is ‘False’. These three points are not collinear. Reason: Form the matrix out of 1
2
1 those three coordinate readings: 0026
0 −1 0 −2 .
2002 Clearly the rank of this matrix is 3. Indeed, it has a 3 × 3 minor
00
0 −1
20 2
0
0 which is nonzero. So the criterion (4b) applies, with n = 4 and k = 3, 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 = 3,
in ‘Fact’ at the end of this solution sheet.
(3) First ﬁgure out the constraint on the deﬁning equation of a codimensi...
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This note was uploaded on 01/12/2014 for the course MATH 660 taught by Professor Yasuyukikachi during the Fall '13 term at Kansas.
 Fall '13
 YasuyukiKachi
 Geometry

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