quiz x solutions

# Then the outcome is a 0 2a 1 7 a0

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## This note was uploaded on 01/12/2014 for the course MATH 660 taught by Professor Yasuyukikachi during the Fall '13 term at Kansas.

Ask a homework question - tutors are online