Question

Q1 17 Points Let T: M2x2(R) P2(R), H (2a +b)x2 + (6 – c)x +(c – 3d). Let B = (16 0) (0 :), (1 o) 9)) = (6 7')(*: -) ) 6 :-)) B' = C = (x2,æ, 1) C'= (x + 2, x + 3, x2 – 2x – 6). You may assume that all of the above are bases for the corresponding vector spaces. Q1.1 2 Points Show that T is linear. Q1.2 9 Points Compute [T), (2pt) = Enter your answer here (1 M2x2 (R)]B. (1pt) = Enter your answer here 1P, (R)]E' (3pt) = Enter your answer here (T) (3pt) = Enter your answer here Please write the answers in the following way: (first row separated with commas\second row separated with commasl... Nast row separated with commas) example: This is a 3 x 3 matrix (1,2,3\4,5,6\7,8,9) Q1.3 2 Points Is (T), invertible? (1pt) O Yes O No Q1.4 4 Points 3 -4 Let A= 1) Compute (A]B (2pt) Enter your answer here and (T(A)]c (2pt). Enter your answer here Write both above as row vectors with no blanks. Example: (1,2,3,4)

Answer 1

T: Ma x 2 (R) P₂ (R) given by (2a+b)x² + (6-c) x +(C-3d) a b с d ( LIEC!!).6.30 0 0 Ale have to show that B is Biesis Bore Me x CRJ. LI.(::):):) tly Co o Co + 0 idagpu GR o o (3): D 0 t 688 du CO q dy 0 So X = d = 2 = du=0 i.e all xico

and B is So by des"; B is linearly independent Since dimension of Maxa (R) = 4 Subset of Maxa (R) Conlarning 4 lineanly independent rechons; S. Basis of Maxa (R) 0.6 (i) ! We have to show Shad b is a basis for min B is Led B 2 - 30 in). 1 B + B2 O t B₂ (; 8) + By loo-lo -Bi-2B2 - Bu 0 Bit B t Bu - B2 t B3 B, B₂ +B so B + B₂ t By = 0 ) B, + 2B + By = 0 B₂ B3 - 0 B, B2 t B3 o

Substract ea (i) from eg (i); We have Ba Put The value as Be in ea (i); We have = 0 B=0 Now The value of B2 and B in ea civ); pulting We have; Bi = 0 Now putting the value as B, and Be in ea ci); We have Buco. So all Bi=0; i = 1, 2, 3, 4 Such Shat Basis BIY, + B₂ Y₂ t B₂Y + By Yu = 0 So by den B is linearly independent. co millanly B is a subsel of Mana (R); which is linearly independent and containing y. linearly independent Verdons; hence by deos ' is a Mata (R). Led c = {x?, x, 13 s PCR). We have to c is had xx ² + Vg x tv yil = 0 ; V., V₂, V E R => Y + / 2 X + Yp x² = 0 Comparing colliciend als 1, x, both sicks: basis for P. (R) a

We have ; 8,0 Y, - O C is Linearly independent C is a So by den Ale know Shad dimension of PCR) = 3 ... PCR) - {at but ex²: a, b, c ER} dince dim P (R) =3 and cc P₂ (R) which has 3 lineaelye independent vectors; so basis for POR) Let c'= { x + 2, 313, x 2 x - 6} C P (R) Ale have do show that c is a basis for BCR) het di (x+2) + dz (X+3) + f (x² - 2x-6)=0; s, of, of ER 8, x + 28, + x + 3 % + & x² - 26,2 -68 = 0 > dr. (5, + & 25 ) x + (as, +36-66)... Now Compaining Cofficients We have S=0 Sit - 26₂ = 0 261 +36-66 70

Pulling the value do & in ea 6 and ; We have -- & + 4 =0 26, +362 =0 Now; 2x ea cdi 28, +262 = 0 ex ea cel 28, +362 = 0 Substreach of (a) (a) from cu : s, o So We have all = 0; i = 1,2,3 Hence c'is Linearly independent Since c s P2 (R) ; Containing 3 linearly independent vectors then by deto is a basis fox P. (R) 8-1.1) We have to show that The breanst formation. T which is given by (a b) = (darb) x + (b-c) xalc-3d) is linean

het (a).() с Max2 (R) and & ER 02 b2 w Now T Cene + (2 dz {{:6) TC aita, bitba CitC₂ dit de [a carta,) + (b + by)] x2 + [(b, + b2 - (C, +C₂ )] x + [rita - 30d, ted) Qa, + b ) x + (aa, + bz) x + (b, -C, J x + (6-62 + (6,-3d) + ((-3d2) {20, + b)22 : Cb, C, Dx: (0-3dy] + 2 azt by ) x + (6 - C2 ) x + (6 - 3 d.)} T 2)+7( + ar 02 b2 de C2 bi di be de G bi be TC JAT( a di d

b Now T[(a dded (a b c Mr. (R) La ab Care ad = (ada + ab) x2 + (ab-4c) x + (dc-3ad) = 2 [aat b) x2 + acb-c) x + (c-3d) (C-3d d {(20+b)x2+ Cb -e)x+60-3d) 2.T ( a b Hence by dose of linear deanst. Tis linear Deb: gf T: Vo veo be any freanstormation ; Shen Tis linear its Tre tv) = T(u) +T(v) NuVEV Tcau)= T(u) V U EV and 2 EF

Bob Max (R) to the basis c ok P. (R) 9-1.2) Ale have to find [T], which is The madrix Representation of T from basis Zus -18) hel 23 Deven B-{z, to 8), 2-6 !). -6:) C= {x?, n, 1). Now TCZ) = T (68) 2 x ² = 2.2 to.n the oil T(23)= T(..) x² + x = 1. x² TCZ) = T(88) = -x + 1 = 0.x² + (-1). x tl. 1 T (ZW) = T( 8 ) 1. x² + lo x tool and 2. -3 = 0.x² +0.x+(-3). /

Hence the matrix. Repr. of I werd B and C 0 0 J 3x4 ie [to (2,1,0,01 0,1-01 0.0.1, -3)

. where BE Smillanely : We can kind (TJ {3-4 :').4-(,). %:( ).6=665) and c = (x+2, x3, x ²27-6) C! :) = (2-1) x² + x-3 row T(9) = T x²+x-3. - 6 (x+2) + (-3) (x + 3) + 1. (x²-2x-6) T(Y) = T 2 = (2-2) x² + (- 2 + 1) x + (-1+3) = -x + 2 - 5.C 2+2) 44.(X+3)+ 0.(XP22-6) T(9₂) = T (oo) = -x + (1-3) = -x-2 = (-1) (x + 2) + 0.(x+3) +0. (x-2x-8)

TC 34) = T Cool (2+1) x2 + (-4) = 3x²-x = (-3) (x+2) et 8. (x + 3) + (x-2x-6) hao So [7] 6 -5 -1 -3 4 8 O 3 3x4 -5, -1, -3 1-3 -3, 4, 0,8 l 110,0,3

هیمال meefreix Xnxm Q-1:3) We know Con] is invertible 0.-/-3) Wel know any is invertible iff Rank of X = Shad matrix any Xnxm = [c, - - - Com it all Ci are linearly independent and also we know shad Rank of X Number of independent Column So here [TJoc is invertible if (T) = 4 [T] Rank ds Min (nm) linearly 는 Rank of but Rani do <3 Xnxm d' So Rank do TJ & 4. not linearly independent So [1] is not invertible ire are all Column's C'

Q.1.4) Led A- (84) Led [A] = (0 - 1) +2 (1) +(99)-(. Lit& tay -21-2₂-du - data Linda taz 8 So dit d t du = 3 -di-24 - Lu=-4 - L₂ + 2 = 0 x, - 2 + z = 1 2 w cul ea (8) t ea -2 - -1 -> 12 = 1 Pak She vale ou & in ea ②; We have h = 0, = 1 From ea W We have & = 1 + 2 - 2 =lt1-/ . L=1

From ea (i); We have Xu = 3-21-22 - 3-2 - du = 1 So [A] 1 - 1, 11) TCA)= T (34) (6-4) x2 + (-4) x + (-3) -ux-3 2x2 = dicata) + (x+3) + f (x²-24-6) (Say) -> 2x2-47-) = (x + 2)2 4724,436 + 2 x² (j) So L = 2 L+2 - 299 = -4 2x +34 - 6 t = -3 Put & = g in egn All We have &* + d 2x, * +36 = 9

Now 2 x ea ) 2x + 24 = 0 2x +30 = 9 (-) Subs. (-) (-) -42* = -9 2) 2 * = 9 So l* = -9. So [TCA)] cl (9, 9.2)