Question

Worksheet on Linear Transformations For this worksheet, let 1 • T:R3 R2 be the function defined by T | 2:02 - 13 -I1 + 12 +3.3 • Rezers ir turi imelo (.)- • U-2 - RP su im datelo (1)-(2353" • U:R3 → R2 be the function defined by U 31 + 22-1) 2:02 - 3.01 - 22 • F:R2 R4 be the function defined by F 2:01 Answer the following questions. 1. Post do what ( )" (C. 1.--(-:) 2. Which of the above functions are linear transformations? Explain how you determined your answers.

Can someone please answer these questions. Please show your work/ explain if needed and neatly and visibly. I'll give a rate if its the answer I'm looking for. Thank you in advance!

Answer 1

2. & GdFIR Given, T:IRIR T (2, 2) = (2x2-x) + x2 + 3x), U: IRRIR U nn) = (14+ 22-1, x2-x2) and FIRMIR", F (1, 2) = (2+2 14,34-22, 2x - 3). 1. T(1-1,2)= (2.(-1)-2, -1-1+3.2) = (-4,4), UG-1,2)=(1-1-1, +4²2)= (-1,-1) and F (1-1) = .. (2.(-1)-1, 3,141, 2.1, 1) = (-3, 4, 2, 1) T(CCM,2, Xa) +d (4.927.) () X,X3), (_Y_YA) EIR? = T(CH+dy,, Co2+dY2, Cx, t0z) = (20 22+ 2dym cxy-dyn, - ch-dy, + Ca2+ dy + 3 cx +3d42) = c (222-23, -2 +2 +34) + d (22=%23-Y, TY2 +373) = cT (1 say n) + DT(1/2/3) So, T is a linear transformation. 47€ (4, x,x)+d (vx,x)) = u(cm tdy, cort dyn, Cx+dra) = (Cardy, & Cant dez-1, (cartdy, Jr – ca-den) + (cardy, & cxtdy 2-1, caurdu - cardi) = c (14+ 22-1, au-x)+ d (Y,+Y2-1, 42-4). - c Une) + d (1, 32, 43) Vis not a linear trans formation. Thus,

F (ccm, 742+d(»»)= F (M+dv, CH2+dve) they wa), (1. MJER = (20 x2 + 2 dyq-cx-dy 3cx+3 dy - Cazdys, 2e4+2dy, , -ca-dY2) = c (282-74, 344 -22, 24, -2) + d (2427,3 34, -7₂, 24,,-Y2) =C F (4,0) + d F (1,12) Thus, Fis a linear transformation. since, if Tis a linear transformation from u lov, then, T(CardB) =(TQ) + dT(B) , & B EU 7 where F is the underlying findet 3. {(10,0), (0.1,0), (0,0,1)) is a basis of IR and {(1.0), (0,1)} in a basis of IR ..T(1,99) = (0-1) = 0.(L) -1 (0,1), T(019) = (2,1) - 2.(139) +1(0,1), T(0,01) = (-13) = 7.(1,0) + 3.(0,1). Thus, madrix of Tis: To 2 17 {(10), (o,)} is a basis of IR² and {(2009), (0,1,0,0), (0,0,1,0), (0,0,0,1)} is a basis of IR". ·F (LO) = (-1,3,2,0) = + (1,0,0,0) +3.(0.1,99)+ 2.(0,01,0) +0:(0,001), F(0, 1) = (2,-1,0,-1) = 2,(1,0,0,0) -1.(0,1,0,0) +0.(0,0,1,0)-1(0,0,0,1)

Thus, matrix of Fis: imao 2TOT 4. Let to = (b, bz) EIR" and (9,92,93) Eir? such that T(a,, azjaz) = (6, 3bc) » (2az-az, -a, +92 +3a3) = (bisbal > 2a2-a3=b, , -a,+Q₂ +3 ag=b2 het aa=c, then, az=22-b, fe ay = +3(24-611-62 = 7c- 6361 +62) for CEIR. Thus, for any CER T(7c-(36, 762), C, 2c-by) = (6, 62), So, Tis onto. 5. Let c=?, bi= 0 and bz=1 in 14 Ther, T(13,2,4)= (0,1) Also, let c=1, b, ao and baal in (a) Ther, T (6, 1,2)=(0,1) So, (13, 34) & (6,1,2) but T (13,2,4)= T (6,1,2). So, Tis not one-one.

6. FOT (7): IR?>IR" defined by FOTO)=FoT(~,923 m) =F(202-; -+xz+34) = (-24, 72 427 6x2 - 2x2+2, E2-3 *z +44 -22-322,482-22, 14-92-3*) = (-2x +7+3, x4 +542-622,422-223, 24-22 - 323) Nows Fot (CCM, w, x) +d(14x)) = FOT(CMtdy, Contdyny cas dry] -4-224-207, +7673+7dy,c*4+dy,+504 +5d92-6c8-Gdya) 4 6x2 + 4 dyz-2025-2dy, C24dy-542-d72-3075-3d43) = c(+244 +3,14*52-603, 40.7203, 4-3mg) +d (-24,+7Y3, Yit 512-619,442+243 14-42-33) =C FOT (1, 22, 23) + d FoT (Y,lasty). So FoT is a linear transformation. Now, like in (3) FOT (1,0,0)= (-2, 1,0,1) = -2.(10,00) + (0,1,0,0) +0.00010)+1(0,0,,1), FOT(0, 1,0) = (0,5, 4,-1) 0.(1,9,0,0) + 5.(0,1,0,0)+4.,010) - 1. (0,0,0,1), FOT(0,0,1)= (7,-6, -2,-3) =7, (1,0,0,0)-6 (0,1,0,0)-2 (0,0,1,0)-3 (00,0,1) Thus, matrix offoT is. F20 77 1 5 -6 O4 -2