I need help with what is not highlighted. 1. Consider the two linear functions h: R3 → P2 and g: P2 → M22 given as here.рр-2g9.H (a + b)x² + (2a + 26)x + cpx + qx + rHсUse these bases for the spaces.(o 2B = (| 1) C = (1+x,1– x, x²) D=3(a) Give the formula composition map go h: R³ → M22 directly from the above definition.Following the definitions gives this.aH (a + b)x² + (2a + 26)x + ca + b2а + 26(a + b) – 2(2a + 26) \a +b2а + 2b— За — 3b(b) Represent h and g with respect to the appropriate bases.Весause0,0,H 2x2 + 4x+11H x2 + 2x +1H Ox2 + 0x +1(1)1we get this representation for h.3/2 1/2)5/2-3/2 -1/2 1/2RepB,c(h) =21Similarly, because(4 ) *- (C )-21+ xH1- xH1this is the representation of g.1-1Repc,D(9) =1/21/3 -1/31

Given- Consider the two linear functions h: ℝ3 → P2 and g: P2 → M2×2 given as here. abc ↦ a + bx2 + 2a + 2bx + c px2 + qx + r ↦ pp-2qq0 To find- Represent h and g with respect to the appropriate bases.Explanation- Firstly, we will represent h with respect to the appropriate bases. Now, h111 = 2x2 + 4x + 1Now,2x2 + 4x + 1 = a1+x + b1-x + cx2 = a+b + a-bx + cx2⇒2x2 + 4x + 1 = a+b + a-bx + cx2By Comparing the coefficients, we geta + b = 1 ......1a - b = 4 ......2c = 2 ........3By solving equation 1 and 2, we geta = 52b = -32∴ we geth1 = 52-322 Now, h011 = x2 + 2x + 1Now,x2 + 2x + 1 = a1+x + b1-x + cx2 = a+b + a-bx + cx2⇒x2 + 2x + 1 = a+b + a-bx + cx2By Comparing the coefficients, we geta + b = 1 ......4a - b = 2 ......5c = 1 ........6By solving equation 4 and 5, we geta = 32b = -12∴ we geth2 = 32-121 Now, h001 = 1Now,1 = a1+x + b1-x + cx2 = a+b + a-bx + cx2⇒1 = a+b + a-bx + cx2By Comparing the coefficients, we geta + b = 1 ......7a - b = 0 ......8c = 0 ........9By solving equation 7 and 8, we geta = 12b = 12∴ we geth3 = 12120 Therefore, we get h = h1h2h3 = 523212-32-1212210⇒h = 523212-32-1212210Now, We will represent h with respect to the appropriate bases. g1 + x = 0-210Let0-210 = a1000 + b0200 + c0030 + d0004 = a2b3c4d⇒0-210 = a2b3c4dBy comparing the coefficients, we geta = 02b = -23c = 14d = 0i.e., we geta = 0b = -1c = 13d = 0So,g1 = 0-1130 Now, g1 - x = 02-10Let02-10 = a1000 + b0200 + c0030 + d0004 = a2b3c4d⇒02-10 = a2b3c4dBy comparing the coefficients, we geta = 02b = 23c = -14d = 0i.e., we geta = 0b = 1c = -13d = 0So,g2 = 01-130 Now, gx2 = 1100Let1100 = a1000 + b0200 + c0030 + d0004 = a2b3c4d⇒1100 = a2b3c4dBy comparing the coefficients, we geta = 12b = 13c = 04d = 0i.e., we geta = 1b = 12c = 0d = 0So,g3 = 11200 Therefore, we get g = g1g2g3 = 001-111213-130000⇒g = 001-111213-130000Answer: h = 523212-32-1212210 g = 001-111213-130000