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Question Number 116746 by bemath last updated on 06/Oct/20
Find an orthogonal matrix A whose  first row is u_1 = ((1/3), (2/3), (2/3)).
FindanorthogonalmatrixAwhosefirstrowisu1=(13,23,23).
Answered by john santu last updated on 06/Oct/20
First step find a nonzero vector   w_1 =(x,y,z) that is orthogonal to u_1 .  for which ⟨u_1 ,w_1 ⟩ = 0=(x/3)+((2y)/3)+((2z)/3)  or x+2y+2z = 0 . One such solution  is w_1 =(0,1,−1). Normalize w_1  to   obtain the second row of A, u_2 =(0,(1/( (√2))),−(1/( (√2))))  Next step find a nonzero vector w_2 =(x,y,z)  that is orthogonal to both u_1  and u_2   for which ⟨u_1 ,w_2 ⟩ = 0 and ⟨u_2 ,w_2 ⟩ =0  ⇒⟨u_1 ,w_2 ⟩ = (x/3)+((2y)/3)+((2z)/3) = 0   ⇒⟨u_2 ,w_2 ⟩ = (y/( (√2))) −(z/( (√2))) = 0  set z=−1 and find the solution   w_2  = (4,−1,−1) . Normalize w_2   and obtain the third row of A , that is  u_3  = ((4/( (√(18)))), −(1/( (√(18)))), −(1/( (√(18)))))   Thus A =  ((((1/3)          (2/3)            (2/3))),(( 0           (1/( (√2)))        −(1/( (√2))) )),(( (4/( (√(18))))  −(1/( (√(18))))  −(1/( (√(18)))))) )   We emphasize that the above matrix A  is not unique.
Firststepfindanonzerovectorw1=(x,y,z)thatisorthogonaltou1.forwhichu1,w1=0=x3+2y3+2z3orx+2y+2z=0.Onesuchsolutionisw1=(0,1,1).Normalizew1toobtainthesecondrowofA,u2=(0,12,12)Nextstepfindanonzerovectorw2=(x,y,z)thatisorthogonaltobothu1andu2forwhichu1,w2=0andu2,w2=0u1,w2=x3+2y3+2z3=0u2,w2=y2z2=0setz=1andfindthesolutionw2=(4,1,1).Normalizew2andobtainthethirdrowofA,thatisu3=(418,118,118)ThusA=(13232301212418118118)WeemphasizethattheabovematrixAisnotunique.
Commented by bemath last updated on 06/Oct/20
waw... thanks a lot mr farmer math
wawthanksalotmrfarmermath

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