Question

linear algebra help.

Let A = 10 1 0 A.) Show s, is LI and forms a basis of R. - Dے واک وه ک ( ه ا ك (.) ( 2/ 5 24 (5) رک) pan 3 | 2 | - |

S2=AV1,AV2,AV3)

Answer 1

FI 0 07 Let A = 1o. l ol Si={ vi, ve, Us} whele Vi=107 V TO Fil V2 = 10 . (A) Let ai, az and q be the three scalalf such that qutqV2 + Q3 V3 =0 1 2018-03-09-08 Tata27 101 [ 927 93 ] [0 equating the collesponding co-ordinates, we have a +93 = 0 - 1) az 30 - 12) Czta3 = 0-13)

(2) from equations (2) and (3), we have Cla= 0, 0+43 = 0 =9330 from equation (I), we have anto 30 9.30 We have a=0, a zo, a =0 Since all the three scalalt 41, 42 and 93 ale gulo, telefore the vectors . Vi va and Vz ale lineally independent, in eo the set Si={v, vs, Va} is lineally inclebendout luhete v = To] v Foc Now every subset of the vector space . Rn having holineally independent vectots is a basis of the nodimensional vectol Space Rh. Hence the set Si=ENY V } having

(3) three linecally independent vectors is a basis of the vector Space R² (B) Sa= favo, Ave, Avs? Now AV, Il lolo Fitoto1 fil. Joitotol = 10 Titoto il Ti 007 107 Fototo] Ita= 0 = 0+1 to Ti I 01 lotto Anti 0 OJTAT Fito+07 TO 1 0 || 0 ototo LL OLI] Litoto V ... Avg = 1

(4) Pri] 1907 A ? LJ, LJ, L Let by ba and by be the three scalabs such that bilavi) + ba (Avs) + b3 14V3)=0 J Litir in To 0 =10 oltb t bi tba 1 ba = . [bit bat b3 ) L equating the Collesponding co-ordinates We have e bit 63=0 w 4 bo=0 - 15) bitbatb3=0 - 6). feom (s) and (6); we have bit 63 30 -17) Now equations (4) and (7) ale same. so we have bitb3=0 Let b3 be arbitealy and equal to la a from (7), bitl=0 : b3=13

(5) Ź bi = -1 we have bi=-7, ba zo. and b3 = 1 Since call the thifce scalabs bi, ba and b3 cole not equal to zero, the lefole the vectold AVI, AU and AUz ale clineally dependent and hence the set Sa={AVI, AVs, AU3) is dincally dependente (C) Now let aloka and ofs be the thece scalats such that R = Gil Avi) + 01 (Ava) + 33 (AV3) T17 7 at altas 7 Ole Lolitatul J

(6) equating the corresponding Co-ordinates We have Git %3=1 417) 92 = -18) att 3 =4 (9) flom (8) and (9), we have alitat24 auto=2 lloj flory (7) and 110), we have the light hand side of the same equations are different, therefore the system of equations (7), 18) and 19) is not consistent on Hence we have the vector U does not belongs to span (Sz), ie. span { Avo, AV2, AV3} le & Shaun (Sa).

(d) Let Y, Y, Y3 be the three scalaus such that TH 21=yout Savat Yz Uz - - O > St Ys7 I ya = |_ Yat 93 equating the corresponding Co-ordinates, we have gity=1 Ho Satyz = 4 +13) from (13) and (13), we have . QtY3=4 = 43 =2 from (1l), ,+2=1 = 9,=1-2=4 we have y, z), a za and Y3=2 Hence the system of equations (11), 112) and 113) is consistenta

.. we have 3 = (-1) Vita) Vg +1) V3 14 Hence the vectol is expressed [4] as a linear Combination of the vectors Vi, Va and Uz of the set Si, therefore we have 1 € 20 (1)