Consider an invertible linear transformation T(x) = Ax from Rm to R \ with inverse L = T~[ from R" to Rm. In Exercise 2.2.29 we show that L is a linear transformation, so that L(y) = By for some m x n matrix B. Use the equations BA = In and AB = Im to show thatn = m. (Hint: Think about the number of solutions^ the linear systems Ax 0 and By = 5.)

' ntlfi6r fb (-q, hr,,o lr,*niu, Mgtx' &xtt \s X-aYtt r.€iro) oFy \.[_z F.ltnor Aillt is Y -q,xls b2+ c olz = bQ = A2- (1. n.rrns Er,r,rpsp Definitiont-1.. ellipsisthesetofal,loi.nPs, {r,y},suehho.t the sofn thedi.stancestweetwopo'i,ncalled,efociof thelli,psconstant. The rnajor axisthe ari,s througfoci,. The rninor axastheo.ri,sraugth,centererpend,i,etothemajor *ri,s. The verticeofheelli,pse theointwhere thelli,'intersectsmajararis. 2.EeuarroNs (*-hl)'-ik)- k)':,1|(*-hh)z ,(y-k)'_,, n -(a-V---;;-^ o' -(y-a'-