Question

Solve #2 & #5

Type Q. Explain why the following sets of vectors are not bases for the indicated vector spaces. 0 {(2, 3), (4,7), (8,17)} for (R² @ {x²+2x+1, 7x²x+6} for IP2 * {[14],[14]} for Merz @{(1,0,0), (0,2,0), (0,1,1),7,,87} for © {sinx, xx} for F ©{(0,0,0), (1, 2, 3), (7,1, 53 )} for R?

Answer 1

Date Scalars) Consider &(x²+2x+1+B (7x²-a +6=0 (2, 3 -> 0x2 +232 +Q+782?-BX +56= 0x270.+0 (Q+782? +20-B)2 + (+63) = 0.42 to 2 te By comparing we get & +78=0 a 20-8=0 66 Q+63=0 J Consider augmented matrise, I 7 olor 2-1 o io =[AB] apply R2R, Ra-R, 1 1 7 olo To-15 olo Lotolo sophy Bastille Ti 0! 6) apply 15 R₃-R2 after that, RR R we get oloo = 290k( A B =rank (A) stampo = No. of unknowns. i system 6 has only one solution ie, q= B = 0 . Vectors are linearly independente But, we can't get linear spank x + 2x+1, 7x² x t6} = 1/₂ we can't express 8x²+x+8 as a linear Combination of x²+2x+1, 7x²-x+6. to for somx ret f. = sinx, fza x² Elf f2 - Sinx 22 = 2xsinx-x²cosx Iti cosx 2x f f linearly independent. But we can't express every funn as a of sinx & x² p.g. tanx. i limear span of sing, x2 F F linear combination