quiz x solutions

Namely suppose a0 x 0 a1 x 1 a2 x 2 a3 x 3 a4 x 4

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Unformatted text preview: on 1 hyperplane that contains those three points. Namely, suppose A0 X 0 + A1 X 1 + A2 X 2 + A3 X 3 + A4 X 4 = 0 is the defining equation of an arbitrary codimension 1 hyperplane that contains the given three points. Substitute each of X0 : X1 : X2 : X3 : X4 = 1:0:0:2:6, X0 : X1 : X2 : X3 : X4 = 2 : 0 : −1 : 0 : −2 , X0 : X1 : X2 : X3 : X4 = 1:2:0:0:2, into each of the above equations. Then the outcome is 3 and ( ∗) + 2 A 3 + 6 A 4 = 0, A0 2 A0 − 2 A 4 = 0, − A2 A0 + 2 A1 + 2 A 4 = 0. You can solve (∗) easily. First, substitute A0 = 0, and A4 = 1 into (∗). Then A3 = −3, A2 = −2, A1 = −1. Substitute these data back into A0 X0 + A1 X1 + A2 X2 + A3 X3 + A4 X4 = 0. So − X1 − 2 X2 − 3 X3 + X4 = 0. Or, equivalently, X1 + 2 X2 + 3 X3 − X4 = 0. Second, substitute A0 = 1, and A4 = 0 into (∗). Then A3 = −1 , 2 A2 = 2, A1 = −1 . 2 Substitute these data back into A0 X0 + A1 X1 + A2 X2 + A3 X3 + A4 X4 = 0. So X0 − 1 X1 + 2 X2 − 2 1 X3 = 0. 2 Or, equivalently, 2 X0 − X1 + 4 X2 − X3 = 0. 4 So, to conclude, we got the equations of two codimension 1 hypersurfaces both of which contain the given three poin...
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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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