As a conic section in x 1 y 1 without cross term when

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as a conic section inx1, y1without cross-termwhenxy=Px1y1.1.point (0,0)correct2.hyperbolax21+ 6y21= 13.ellipsex21+ 6y21= 14.hyperbolax216y21=15.straight linesx21+ 6y21= 06.ellipsex216y21=1Explanation:The quadratic relation2x24xy+ 5y2= 0can be written in matrix terms asxTAx=xT2225x= 0wherex=xy.To eliminate the cross-term we orthogo-nally diagonalizeAby finding the eigenvalues
Version 047 – NewEXAM04 – gilbert – (53525)5λ1,λ2and corresponding normalized eigen-vectorsv1,v2ofA.For thenA=PDPTwithP= [v1v2],D=λ100λ2,andxTAx=yTDy=λ1x21+λ2y21,settingPy=x,y=x1y1,x=xy.SinceAis symmetric,Pwill be orthogonal ifλ1̸=λ2. Butdet[AλI] = (2λ)(5λ)4=λ27λ+ 6 = (λ1)(λ6) = 0.Thusλ1= 1 andλ2= 6, so that inx1, y1coordinates the quadratic relation becomesx21+ 6y21= 0,the only solution of which is the point (0,0).00910.0 pointsDetermine the singular valueσ1for the ma-trixA=200232.1.σ1=192.σ1=17correct3.σ1= 254.σ1= 325.σ1=21Explanation:By definition,σis a singular value ofAwhenσ=λandλis an eigenvalue ofATA;σ1is the largest of these singular values.NowATA=203022200232=13668.But thendet(ATAλI) =13λ668λ=λ221λ+ 68 = (λ17)(λ4).Consequently,σ1=17.01010.0 pointsWriteyas the sum of a vector in Span{u}and a vector orthogonal touwheny=424,u=010.1. y= 2010+ 4101correct2. y=000+4243. y=2010+ 41114. y=12010+128585. y=12010+12838Explanation:
Version 047 – NewEXAM04 – gilbert – (53525)6Let us writeyasy= projuy+zwhereprojuyis a vector in the subspace spanned byuandzis orthogonal tou. Thenprojuy=y·uu·uu

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