A basis for this line is the vector 1 1 0 22 p is the

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1, 1) = 0 and (a, b, c) · (1, 1, −1) = 0. That is all simultaneous solutions to a + b + c = 0 and a + b − c = 0, a line. A basis for this line is the vector (−1, 1, 0). 22) P is the nullspace of the matrix [1 1 1 1]. We can read off a basis for P ⊥ , it is the normal 1 1 vector, namely . 1 1 26) Pick two perpendicular vectors and then take their cross product to get the third. So 1 3 1 4 works. Now A = 1 −2 1 −1 −5 1 1 1 1 3 1 300 4 = 0 14 0 . AT A = 3 −2 −1 1 −2 1 4 −5 1...
View Full Document

This note was uploaded on 01/30/2014 for the course MATH 2940 taught by Professor Hui during the Fall '05 term at Cornell.

Ask a homework question - tutors are online