quiz x solutions

Then the outcome is a 0 2a 1 7 a0

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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 non-zero. 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.

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