Chapter 2ISM:Linear AlgebraChapter 22.11. Not a linear transformation, sincey2=x2+ 2 is not linear in our sense.2. Linear, with matrix0200031003. Not linear, sincey2=x1x3is nonlinear.4.A=93-32-914-9-25155. By Fact 2.1.2, the three columns of the 2×3 matrixAareT(e1), T(e2), andT(e3), sothatA=76-1311917.6. Note thatx1123+x2456=142536x1x2, so thatTis indeed linear, with matrix142536.7. Note thatx1v1+· · ·+xmvm= [v1. . . vm]x1· · ·xm, so thatTis indeed linear, with matrix[v1v2· · ·vm].8. Reducing the systemx1+ 7x2=y13x1+ 20x2=y2, we obtainx1=-20y1+7y2x2=3y1-y2.9. We have to attempt to solve the equationy1y2=2369x1x2forx1andx2. Reducingthe system2x1+3x2=y16x1+9x2=y2we obtainx1+ 1.5x2= 0.5y10=-3y1+y2.No unique solution (x1, x2) can be found for a given (y1, y2); the matrix is noninvertible.52
ISM:Linear AlgebraSection 2.110. We have to attempt to solve the equationy1y2=1249x1x2forx1andx2. Reducingthe systemx1+2x2=y14x1+9x2=y2we find thatx1=9y1+2y2x2=-4y1+y2orx1x2=9-2-41y1y2.The inverse matrix is9-2-41.11. We have to attempt to solve the equationy1y2=1239x1x2forx1andx2. Reducingthe systemx1+2x2=y13x1+9x2=y2we find thatx1=3y1-23y2x2=-y1+13y2.Theinverse matrix is3-23-113.12. Reducing the systemx1+kx2=y1x2=y2we find thatx1=y1-ky2x2=y2.Theinverse matrix is1-k01.13. a. First suppose thata= 0. We have to attempt to solve the equationy1y2=abcdx1x2forx1andx2.ax1+bx2=y1cx1+dx2=y2÷a→x1+bax2=1ay1cx1+dx2=y2-c(I)→x1+bax2=1ay1(d-bca)x2=-cay1+y2→x1+bax2=1ay1(ad-bca)x2=-cay1+y2We can solve this system forx1andx2if (and only if)ad-bc= 0, as claimed.Ifa= 0, then we have to consider the systembx2=y1cx1+dx2=y2swap :I↔IIcx1+dx2=y2bx2=y1We can solve forx1andx2provided that bothbandcare nonzero, that is ifbc= 0.Sincea= 0, this means thatad-bc= 0, as claimed.53
Chapter 2ISM:Linear Algebrab. First suppose thatad-bc= 0 anda= 0. LetD=ad-bcfor simplicity. We continueour work in part (a):x1+bax2=1ay1Dax2=-cay1+y2·aD→x1+bax2=1ay1x2=-cDy1+aDy2-ba(II)→x1=(1a+bcaD)y1-bDy2x2=-cDy1+aDy2x1=dDy1-bDy2x2=-cDy1+aDy2(Note that1a+bcaD=D+bcaD=adaD=dD.)It follows thatabcd-1=1ad-bcd-b-ca, as claimed. Ifad-bc= 0 anda= 0,then we have to solve the systemcx1+dx2=y2bx2=y1÷c÷bx1+dcx2=1cy2x2=1by1-dc(II)x1=-dbcy1+1cy2x2=1by1It follows thatabcd-1=-dbc1c1b0=1ad-bcd-b-ca(recall thata= 0), asclaimed.