Question

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50. In Parts (a)-(b), you are given a pair of ordered bases B and B' for R2. Find the change of coordinate matrix that changes B'-coordinates into B-coordinates. (a) B = {(1,3), (2,5)} (b) B = {(1,0), (0,1)} and and B' = {(1,0), (0,1)} B' = {(1,3), (2,5)} ) is the change of 51. Let B = {(1,1), (1,0)} and let B' be an unknown basis for R2. Given that Q = ( coordinate matrix that changes B'-coordinates into B-coordinates, find (1,7)]B.. 52. Let T : R3 R3 be the linear transformation that projects every vector in R3, originating from (0,0,0), onto the plane x + y – 2z = 0 (see illustration to the right). Let B be the standard basis in R", and let B' = {(1,1, -2), (-1,1,0), (2,0,1)}. Throughout this prob- lem, you may use the facts that (1,1,-2) is orthogonal to the plane, while the other two vectors in B' are contained within the plane. TW) (a) Use symmetry to calculate (T]. (b) Calculate the change of coordinate matrix, Q, for taking B'-coordinates to B-coordinates. (c) Use Theorem 2.23 to find (T6, and then use this matrix to find a formula for T. Feel free to use a computer or calculator to find Q”.

Answer 1

(5 @ 8= {(1,3), (2, 517 and ß' = {(1,0), 10,157 T[(1,0)= a (1,3) + B (2,5) T[(1,01] = 12,34 ) + (28, 5B) (10)] = (x+2/3 , 34+58) => 0423=) ; 30+5g=0 - © 3x cen 0 - len ③ 34+ 68=3 3 + sp=o By egr x+ 2x3 =) [x=-5) T[(1,0)] = -5(1,3) + 3 (2,5) Noro T [10,90] = 2 (1,3)+ B12,5) TE10,1)] = (0,3%) + (28,58) T[to,s] = (2+2ß, 34+58) 2) 2+28 =0 - ; 32+5821 w 3xejn ③ - egr G So Co-ordinate matrix that 32+6B=0 32 + 58=1 I changes ß - coordinates into |ß- Coordhates By egh ③ x-220 ros 27 18=2 T(0,= 2(13) =(2,5)

8 B = {1,01, (0,177 and ß = {(1,3), 12,51 4. T[ 11,37] = (do T(1,3)] = (1,3)] = 2(1,0)7 810,1) (0,0) +10.ß) T(1,3)] = (248) 2) Q=1 3=3 T [11,37] = 1 (1,01+ 310,1) Now T [12,50] = x 11,014 Blo,1) T[12,51] = (0,01)+10,B) T[ (2,5)] = (2,8) 2) X=2 ;8=5 T[(2,5)) = 2(110)+ 5(0,1) So change of Coordinate matrix that changes B'- coordinate into ß- Coordinate - T 27 3 su 0 B= 24,, (,0)) : (- (37) het B = {(a,b), (ad) Tlab) = 2(1, 1) (10) Tlab) = 12,2)+(-1,0) 2) Tlab) = (2-), 2to) Tlaib) = (1,2)

: 20-33=7 - T(Cd) =-3 (1.1) +4(10) Ted) = (-3,-3) +14,0) Tead) = (1-3) B = 4 (1,2), (1, -3) Q+Bz) - 0 2x & " (1) - Can ② 24+28=2 22-38=1 SB=-5 (B-1) By lez 0 2-1-12 =2) (117) = 2 (112) + B (1-3) (117) (0, 22) + (8,-38) (117)= (x+8, 22-38) So (1,7) = 2(1,2)-1 (1-3)