Question

MAT 2113 Test 4 Sp20 Name You must show all work to receive full credit. 1. Find the coordinate matrix of x relative to the nonstandard basis for r. B-11.4.6.6.11.01.17.0,101), x(3, 19, 2) 2. Find the transition matrix from to. 3. Find the transition matrix from B to B. B - 1.1.1.1.1.01.1,0.01) B-1.2.3.0.1.02.(1,0,1)) 4. For u =(-1, 1, 2) and y = (-2,3,1), find: a) M b) unit vector in the direction of v c) d) d(u,v) e) proj 5. For u =(4,1, -3) and y=(0,3,1), and ) **, *2*, find: b) d(u,v)

Answer 1

1. B' = q (1, 4, 6), (8, 11,0), (7,0,10)) X =(3, 19,2) (3,19, 2) = 4(1,4,6) + 42 (8,1190) tez (7,0,10) e, C C are arbitrary where constant => (3, 19, 2) = (a + 8C24 763 , A G + 11C23 bet103)

G+82 +72 73 o 4G+ 11 C2 = 19 C₂ = 19-49 - 64 +10 G 22 7 34+563 21. 5 4 2 1-34 from ☺ - 4+ 8 (19-44) 761 > llat 152-32 G +7(1-3) = 3. 27 152-21G 7-219 - 2 3 S. 27 760-1054 + 77-2319 55 27 837 - 3369 = 165.1 37 672 - 3364 3) a = 2

و ها (2) - 2 در را در ر2 = C زهراار%) 1 + (6ر14) زارا (7) (1) - = (2 ا ارد) و coordinate matter of V. to the norstanderd bases the o relatice د / .9 ازهر ۱٫۵ (۱٬۱ , ۰ ), (ارا )) اداره را) دره رار),(3ر2(۱) ( = ام ز هرا) (0 ,0 ,0) ر) + (3 ,2 ,1) = ( 1 / 1 ) از دور + ع2 )+ ) = (۱ را (1) رام 2 CFC = 2 f22' 1 - 3= ع ہو (= = 3- ) + 2 = 232/ 2 =0 0 = 2)

(1,1,1) ? (1,2,)+1(0,1,0) 41 (1,0,1). 0,1,0) = 04 (1,2,3) + d2 (0,1,0). tdz (1,0,1) > (1,1,0) = (de td3 , 2d7 +d, & 301 + 3) de + d₂ = 1 1 2d, td 22 11 34 +d300 >) 7-30421 | 220 > dz 2-3d4 2) - 2de zi - 26-2) + d₂ ²1 ) d3 - 3 2) dy = -1 1 = 1 + d2 = 1 =). d2 2.2. 1. (1,1,0)=-1 / 2 (1, 2, 3) + 2(0,1,0) + Ź (1,0,1). - Now, (1,0,0)= h, (1, 2, 3) the (0,1,0) tha (1,0,1) > (1,0,0) alaituz, 2h, th, 341th₂) hitha 11 anith 20 13th 26 2) h, -34,21 | 23-24 h 2-3h, 39,-121 onza

(1,0,0) = 6) (1,2,3)+1(0,1,0)+261,0 & matrie į To - 1 1 2 1 4. U=(-1, 1, 2) v = (-2,3,1). a) Iv 12 = vw=(-2,3,1). (-2,3,1) +12 = (-2,2 +3 = 4tati = 14 . Ivl=W14. b) mit vector of a w IVI c) U.V = -1,1,2).(-2,3,1) 22+ 3+2 77

aux vd u = (-1,1,2) V= (-2,3, V dlu, uz llu-vll hat Now, u-va (1,-2, 1). 11u-vil 12+(-28²+12=6. o so llu-vil = 56. e proj ua luov) Tv2 v. 14. v) 2 (-1,1,2).(-2, 3, 1). = 2 +3+2 z 7 IV1= (2) + 32 +2 9.4+9+1 2 14. So, proj u = 7 (-2, 3, 1) }(-2,3,1)

5. uo (4,1-3) v= 10, 3, 1) v) = 2y Vitu,v +24₂3 a) <u, v7 = 24480 ) + (1X3)+2 (-3)*1) = 280 + 3 - 2(341) 30+3–(2X3) 20+3-6 -3. 6) d (la, 4) = 11u-vil u-v=(4,-2,.-4) ||u-v11 2 = (4)² + (-2)2 + (-982 23244 236 i i 11u-vil a 3 a6. & d (u, v) =6 .