Question

(1 point) Consider the ordered bases B = {-(7 + 3x), –(2+ x)} and C = {2,3 + x} for the vector space P2. a. Find the transition matrix from C to the standard ordered basis E = {1,x}. TE = b. Find the transition matrix from B to E. Te = c. Find the transition matrix from E to B. 100 TB = d. Find the transition matrix from C to B. TB = 11. !!! e. Find the coordinates of p(x) = 3x – 3 in the ordered basis B. [p(x)]b = f. Find the coordinates of g(x) in the ordered basis B if the coordinate vector of g(x) in C is [9(x)]c = [q(x)]b =

Answer 1

Solution. Given th Given that B-{-47138), -(24x)} (= {2, 3+x} (a) E = {1,83 2 = 9 (1) +(268) = ,= 2,69 -0 .. 2 =201)+ (0)(1) T= $2 27 (b) TEE -4+3x) = -1-3 = -761)+(-3)(x) -(2+36) = -2- = -2(1)+ (-1)() (c) TEB 1= 9697-32) +(21-2-2) >> -74-262=1 [:Equating the constants on both sides] And -34 -2-0 Equating the coefficients of a on both sides y (2=-39 -16 - 26-3G) = 1 -74 +60=1 -16 = 1 -> CR-1 (2 = -3(0) = -3(-1)-3 C -1, (2-3 Now, 1 -14-7-3x)+3(-2-x) 06 - (3 (-7-3x) +C4 (-2-2) - 3'3-C4 =1 [:: Equating the coefficients of Ic' on both sides] And -713 - 264 = 0 [.; Equating the coefficients > C4 = 763 - 363 + 16 =1 -663 +763=2 (3=2 ::C4 =-113 -72) --1 (3-2, (4--7

(d) TB 2+20) = 2,(-7-326) tag (-2-2) 1 - 79,- 292= 2 . Equating constants] And -3a, - a2 = 0 or Equating coefficients of x'] → 22 - - 3a, - 79, - 26-39,)=2 > -79, +62,=2 -,= 2 = 9,--2 -3(-2)-a, = 0 → 22 = 6 .: 9, 3-2, a, -6. 3+ x = b, (-7-37) +62 (-2-3) 4452 - -Tbi - 262=3 ² -76,-2b2 = 3 -3b -b2=1 * x2 -66,- 2b2 = 2 bera - b = -3 bi-b2 = 1 - 6=-1 -3(-1)762= 1 → b2 = 3 ibiz-1, 62=3 TB = (2 -17 (e) PCX - 34-3 33-3 = (-7-3x)+(2(-2-1) -16 -202 = -3 , -16-21;= -3 -36, -2 = 3 ->425-6622 C -9 -34,- (= 3 -31-9) - G = 3 +21-3=6₂ - 2 - 24 : 612-9, (23: 24 :

(5) - °[sy) -: 91 – “о '9-= 'ь"? 91 – “о: г-ы+ = “о - 7="p- (9-) – е 7- "о- 'bg - 9-= p -а, - (-че (+) в) – 60, - 24, 24 -а, - 20, = lo ч. , . - 2 -За – а, - 10, - 2а, — to 2x +10 - о, (-1-21) +0, (-2-) + 7 = 1.7 +2 тт = (+) + (1) (1) - (7Ь 5 (-) °C+1) -