Sulotion:
Given that nebce that to the \(\displaystyle{T}:{P}_{{2}}\rightarrow{P}_{{3}}{T}{\left({f}\right)}={3}{x}^{{2}}{f}'\)
a) \(\displaystyle{B}=\le{f}{t}{\left\lbrace{1},{x},{x}^{{2}}{r}{i}{g}{h}{t}\right\rbrace},\gamma=\le{f}{t}{\left\lbrace{1},{x},{x}^{{2}},{x}^{{3}}{r}{i}{g}{h}{t}\right\rbrace}{N}{S}{K}{T}{\left({1}\right)}={0.1}+{0}.{x}+{0}.{x}^{{2}}+{0}.{x}^{{3}}{N}{S}{K}{T}{\left({x}\right)}={3}{x}^{{2}}{c}{o}{\left.{d}{t}\right.}{1}={0.1}+{0}\cdot{x}+{3}{x}^{{2}}+{0}\cdot{x}^{{3}}{N}{S}{K}{T}{\left({x}^{{2}}\right)}=={3}{x}^{{2}}\cdot{2}{x}={6}{x}^{{3}}={0.1}+{0}.{x}+{0}.{x}^{{2}}+{0}{x}^{{3}}{N}{S}{K}{{\left[{T}\right]}_{{B}}^{{\gamma}}}={A}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{0}&{0}\backslash{0}&{0}&{0}\backslash{0}&{3}&{0}\backslash{0}&{0}&{6}\backslash{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}\)
b. That to that the \(\displaystyle{N}{\left({1}\right)}\cdot{A}{x}={0}{N}{S}{K}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}&{0}&{0}\backslash{0}&{0}&{0}\backslash{0}&{3}&{0}\backslash{0}&{0}&{6}\backslash{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{x}_{{1}}\backslash{x}_{{2}}\backslash{x}_{{3}}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}={b}{e}{g}\in{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{0}\backslash{0}\backslash{0}{e}{n}{d}{\left\lbrace{b}{m}{a}{t}{r}{i}{x}\right\rbrace}{N}{S}{K}{3}{x}^{{2}}={0}{6}{x}^{{3}}={0}{N}{S}{K}{x}_{{2}}={0}\)
Let then the \(\displaystyle{n}{\left({1}\right)}={5}{p}{o}{n}{\mid}\le{f}{t}{\left\lbrace{1},{0},{x}{r}{i}{g}{h}{t}\right\rbrace}\) spon
let \(\displaystyle{P}{\left({x}\right)}={7}+{8}{x}+\gamma{x}^{{2}}\in{P}_{{2}}\)
hence that to
Let then to T(P) = A - \begin{bmatrix} x\\ B \\ \gamma \end{bmatrix} = \begin{bmatrix}0 & 0 & 0 \\ 0 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 6\\ \end{bmatrix} \begin{bmatrix} x \\ 0B\\ \gamma \\ \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 3B \\ 6 \gamma\\ \end{bmatrix}ZSK
Then the \(\displaystyle{R}{\left({T}\right)}={s}{p}{a}{n}\le{f}{t}{\left\lbrace{\left({0},{0},{1},{0}\right)}^{{T}}{\left({0},{0},{0}\right)}^{{T}}{r}{i}{g}{h}{t}\right\rbrace}\) is spon \(\displaystyle\le{f}{t}{\left\lbrace{x}^{{2}},{x}{r}{i}{g}{h}{t}\right\rbrace}\)
c. \(\displaystyle{A}=\le{f}{t}{\left\lbrace{x}^{{-{1}}},{x},{x}^{{2}}+{1}{r}{i}{g}{h}{t}\right\rbrace}{B}=\le{f}{t}{\left\lbrace{x}^{{3}},{x},{x}^{{2}}{r}{i}{g}{h}{t}\right\rbrace}\)
Let the then PSKT(x - 1) = 3 x^2 - 1 = 0.x^3 + 0.x + 3 x^2 + 0.1

T(x) = 3x^2 - (1) = 0.x^3 + 0.1

T(x^2 +1) = 3x^2(2x) = 6.x^3 + 0.x + 0.x^2 + 0.1

[T]_{A}^{B} = \begin{bmatrix}0 & 0 & 6 \\ 0 & 0 & 0\\ 3 & 3 & 0\\ 0 & 0 & 0\\ \end{bmatrix}

T(x) = 3x^2 - (1) = 0.x^3 + 0.1

T(x^2 +1) = 3x^2(2x) = 6.x^3 + 0.x + 0.x^2 + 0.1

[T]_{A}^{B} = \begin{bmatrix}0 & 0 & 6 \\ 0 & 0 & 0\\ 3 & 3 & 0\\ 0 & 0 & 0\\ \end{bmatrix}