Question

2) Let T be a linear transformation from P3(R) to M22(R). Let B= (1+2x + 4x2 + 8x3), (1 + 3x + 5x2 + 10x3), (1 + 4x + 7x2 + 13r%),(1 + 4x + 7x2 + 14x²). Let C= [] [ 1];[1 ] [ ] 0 17 40 Let M= 13 31 36 124 22 52 -61 -209 23 55 -64 -220 be the matrix transformation of T from basis B to C. -47 -161 The closed form of the transformation is T(a + bx + cx2 + d2%) = 3) Let T be a linear transformation from M22(R) to P3(R). BE [11] [1 1 -2 ] [73] [4 -1 0 2 3 3) Let C = (1 + 12 + 0x² + 0x3), (0+ 12 + 1x² + 0x), (0+ 0x + 1x2 + 12'), (0+ 0x + 0x2 + 1x3). -1 1 0 1 -5 -5 5 -7 Let M= be the matrix transformation of T from basis B to C. -9 16 28 13 -23 -13 23 16 The closed form of the transformation is T 7[:]

Answer 1

Its Augmented maerion is 2 а 1 1 - 2 0 0 b R4-2 R32 RBTORI 0 2 -1 2-1 Ctza to T 0 1 0 3 e d-c. 3 -3 A -2 o 0 -1 ㅗ e-ard R2+Ri Ri -(R3TRA). o O c-a-d btc-a-d 1 b I 0 0 0 0 0 0 -1 0 1 0 3 0 0 - - 0 3 letra d-20 C+2a d-20 0 R3+BR2 R4 - R2 D 0 0 0 c-a-d 0 0 0 -1 btc-ad 0 1 0 0 1 3 brac-a-3d 0 otolatad-3c-b -CA Since o Ga@ard tot & Grenad 3 btc-a-d - 3 b4c-a-Bd C 23b74c-a-3d C3 =Bexbra-22 -C3 = a +20-30- cq zatd-boc I is a limearmausping 80 [1] 297[:] +51 [3.93 +9 TC3%9716418:37 ADT [a b] = +(c-a-d) (-+-6-215171223)+(35+4c-a-3d) (1–426 -1497 793) +(36+b-a-2d) (0+5%+187–1003)+ Ca+d-b-c) (1-64-200271003) Or, T [ J.= (2+2b+20 - 1d)+(-a -19924-20)2+(-32-45 -4e + 1 d) +Cath-3d) on; treb] = Ca+25+20-d) +(-6-6-C+2d)x+ (-3a-4b-4cttc) a² + (afb-3d]x3 CS Scanned with CamScanner

Here And matria Thus o Mi of T from B to Ċ TE] = -4(174970*7023) - 540+4 07291 + 003) -16C0+on+1977 293) +28(0:700 +on+ 1 93) on, 15) + 1-GM -2101 71253 and PI10 1.61+ant on ton 3) -56OTALT 127 023) -I-2 910+ on PAMP193) +1600+on+Ott 293) 1 - AM - LAM 7700 TE = 0(ltant on t003) + 51 Otant AntonB) 1300+on+ 2017 4003) - 23604084094113) = 0 + 5x +18m_song [1-3] I litt alt om 1003) -71041014 2012 013) --19 (0+09A113)+29(0+084 on An = 1-6n-2000 --Y 0 5] 2 3 TP2 -1 1093 how let T@ 350 274]*2 [4 1-2399[]+971 la b] = ato-estilu -a-ca 1-29 -q +29-4 -49-2 + 3 C 3 -3 CA ata +264 z a -G-CA a b 29-G+2C₃-C4 = --44-262 +39-304 2d CS Scanned with CamScanner
Answer of 3 in the above page.
Given that TIPER) → M2 CIR) is a limearmapping simee M be the matrix Of T from B6o e so that 2 ) T(4+20+ ant sm) = 131607 +31 [18]*8*[81]-360 5.7-426e. TGART AMB 2 3 3) 운 ] 136 -80 1 simee we know if m = (ai;) myn basis is the matrix of T from any to {ms-Romy then Teaci) => aji @j!!! (2012) j21 T(A+BL+ 5mint 10x) = 47 [6J +10%5-9750:1-161[69] 11 40 -A1 -104 TCatanti tiga) 2 - 13 2 2649]+527:15-6-1987-209588] 52 22 -61 -185 T (AT ANATA = 23 [68]+55C641-64439]-220169] 0 23 55 -64 -143 crow on at but cxdxb Let (atbutes dro) e P₂ (IR) (domain of T) Then we can write at brt chdi as a linear combination of basis vectors of B atbet ant dro = q (A+2AHAP 893) + q (4+3nt. Strona) + (1+44 + 70W+1393) + 4 (14444 70M JAS) z (at G+ G + (4) +(29+39 +4G+ACA) x + 149 +59 +40 +704)t.184 +1067 1966 1967 143 Faluating both sides the coefficients of various Of a... 679 +6+4 za 29 +2 9+4 C₂ +4G zb 40+59+7C3+7C4 20 Scanned wit&GTSE Get 130s +144= power

The augmented matribe of this systemu zl3 a b 2 3 4 4 A 5 77 % 10 13 14 с d Taring now operation a o 6-2a 0 0 0 'a+c-2b 0 01 d-b-c- 2a Be-2b-d 'RA - 2 R3 R34 R3 -2 R2 R2-2RI 1.0 oto crespectively) 0 A 2.2 O- 1-1 c-2b Ri+R3 R2 + R3+R4 Rg-RA 0-1 0-1 2d-2 0 0-10d-2c (res, G a aterry a = arc-2b Y 24 d-b-e-za CA = d-b-c-2a -42-44= 30-26-d -C3 z d-20 G =3c-2b-dt "d-b-c-2a - Ed-20 a 2a+C-26 2 z 2 at3b-20 a 22c-d A = db-c-2a so Since t is a linear meepping from P: CIRI HO MODWR) T Carbat cx dm3) = 9 Tator +4727893) + GT(A+B x + 554 100) + CM (+44 470171375)+44T (4+ 4%+16+ 1493) or, (a+bn4ca7d93) z (a+C-2b) 513 +(2013b-20)717.20 so L-99-10 +(2-0) 2.2 52. tedeb-c-2a) [23 55 21 -36 -61 - 135 - 64 -143 or, Tcar batent dx3] a+ 2b +ord -392-5767406 +3d c2a-9b4C-8d -2a-5b-Bd [Reduired materia CS Scanned with CamScanner