Question

(1 point) Consider the ordered bases B = (1 – X,4 – 3x) and C = (-(3 + 2x), 4x – 2) for the vector space P2[x]. a. Find the transition matrix from C to the standard ordered basis E = (1, x). -3 2 TE = -2 b. Find the transition matrix from B to E. 1 -1 T = 4 -3 c. Find the transition matrix from E to B. -3 1 T = 4/7 -1/7 d. Find the transition matrix from C to B. 17 -5 T? -10 2 e. Find the coordinates of p(x) = x-3 in the ordered basis B. 5 [p(x)]B = f. Find the coordinates of g(x) in the ordered basis B if the coordinate vector of q(x) in C is [9(x)]c = (1,-2). 27 [9(x)]B = -4

Answer 1

a It is given that the ordered bases, B= (1-2, 4-37) C = (-(3+2x), 47-2) and the standard basis E = (1, x) for the vector space 22 [2]. We have to find the transition matrise e to E hit -(3 + 2x). = a.ltb.x and 4x - [email protected] az-3, b = -2, @ = Then the required transition matrix To ed + d-x X .. -2, d=4 il e TE = [-3 - 3 4

Now, we have to find the transition matrix from B to E. ht | - a,.ltb.a and 4-3x = Cilt doox - a, = 1, b =-1, C, = 4, nd,- -3 . To [eo, DJ= [4 =3] = [1] ie. Tot 4 4 -1 -3 Again, we have to find the transition matrix from E to B at i=0₂.(1-x)+ b2 (4-3x) and x= q (1-x) + d2 (4- 3x)

We can write the equations as 1 = CQz+462) - (Az+352) X x=(4+402)-(4+32)x - A₂+462=1, a₂+3b2=0 ( 74 dz = Di G+ Z dz - 1 solving the above equations, Az = -3, b2 = 1 C =-4, dz=1 - T az bz Cz da -4 1 it, Teo [] 1 Now, we have to find the transition matrix from e to B

= + 2 <f - (3+2х) п, (-x) + b (4 - 3x ) 4х – 2 = с (-x) +d, (4-32) - 3- 2x = a +4b, )- (а+зь,)х -2 +4х = (e+4d5) -(c+325)х 2. аз +463 - 3, аз + эb; Cз +4д, = -2 cз+здз = -4 solving the above equations, a = 17, bат - 5, ез = - 10, d3 т° - Ге; ;]" - - - - - - [ 2] - 2 :. Те 2 іс, то B

P(2) aut = 2-3 We have to find now, the coordinates of x-3 in the ondered basis B. Pla) = C. (1-x)+6 (4-30) @, +462)-(6,+363)2 :. C, +462=-3, 6, +362= -4 -1 solwing, C, =s, G= -2 -- [PC]o - [5] Again, we have to find the coordinates of 91>) is the andered bav's B where, [260] = (2-2)

· 9(01--.(3+27) - 2.(42-2) =-3-22-8x + 4 = -10% +1 aut, q (0) -102 +1 = - 3 +4 Cu = C3(1-x) + Cu(4-3x) (Cz+ Cu) x (C₂+464) Cz + 3 C4 = 10 , Cz E sohwing, so 9.eu solving, C₂ = 37, Cu =-9 Thus, the coordinate vector of 9(2) in the ordered basis B is, [907], - (-)]