Question

Problem #8: Find a basis for the orthogonal complement of the subspace of R4 spanned by the following vectors. v1 = (1,-1,4,7), v2 = (2,-1,3,6), v3 = (-1,2,-9, -15) The required basis can be written in the form {(x, y, 1,0), (2,w,0,1)}. Enter the values of x, y, z, and w (in that order) into the answer box below, separated with commas.

Answer 1

Soin om Here we have to find a basis for the orthogonal complement of the subspace of IR4 spanned by the rectors V = (1,-1,4,7), V2 = (2,-1,3,6) , V3 = (-1,2,-9,-15) So let w be subspace where W = span { vi, V2, Vs} The arthogonal complement of w as the null space of the matrix whose rows are the given set of vectors Sponning H. So 2 -1 3 6 - 2 --15J we find the nullspace of A by row reduction & R+R-Ry 11 4 7 1o TRT-2R, TI- 14 2 -1 3 6 10 L 2 -quisto + Rz tR, Lo -5-glol Ritz 1 1 -1 4 7 OR 10 -5 -80 O O O Olo To 1-5 -80 Loo 0 ohol we get 2,- 2,-0,=0 x = 2, tw, 4, -52,-800,=0 Y = 52,+8w, where 24w are free variable So x= (^, Y, Z, (q) = (2,+W,, 52, +8w, Z,, w) = 3,(1,5, 1,0) +W,(1,8,0,1) we get x= (M41 4,12,,(0,0) = 2,(1,5,!,0) +w,(1,8,0, D is solution of Ax=0 so Required Bars can be written in the form 1. B = {(1,5, 1.0), (1,8,0,1)} where x = 1, y =5,2=1, w=8 as bans for orthogonal complement of subspace w. - Above is our required solution