Given T [ x 1 x 2 x 3 ] = [ x 1 x 3 − 2 x 2 − x 3 ] T\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} x_1 \\ x_3 \\ -2x_2-x_3 \end{bmatrix} T ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ x 1 x 3 − 2 x 2 − x 3 ⎦ ⎤
In matirx form we have T [ x 1 x 2 x 3 ] = [ 1 0 0 0 0 1 0 − 2 − 1 ] [ x 1 x 2 x 3 ] = A X T\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} =\begin{bmatrix} 1 &0 &0 \\ 0 & 0& 1 \\ 0 & -2 & -1 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = AX T ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = ⎣ ⎡ 1 0 0 0 0 − 2 0 1 − 1 ⎦ ⎤ ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤ = A X
Where A = [ 1 0 0 0 0 1 0 − 2 − 1 ] A=\begin{bmatrix} 1 &0 &0 \\ 0 & 0& 1 \\ 0 & -2 & -1 \end{bmatrix} A = ⎣ ⎡ 1 0 0 0 0 − 2 0 1 − 1 ⎦ ⎤ ; X = [ x 1 x 2 x 3 ] X = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} X = ⎣ ⎡ x 1 x 2 x 3 ⎦ ⎤
and f ( x ) = − x 3 + 2 f(x) = -x^3+2 f ( x ) = − x 3 + 2
then f ( T ) = f ( A ) = − A 3 + 2 I f(T)= f(A)=-A^3+2I f ( T ) = f ( A ) = − A 3 + 2 I
A 3 = [ 1 0 0 0 2 − 1 0 2 3 ] A^3=\begin{bmatrix} 1 &0 &0 \\ 0 & 2& -1 \\ 0 & 2 & 3 \end{bmatrix} A 3 = ⎣ ⎡ 1 0 0 0 2 2 0 − 1 3 ⎦ ⎤ , 2 I = 2 [ 1 0 0 0 1 0 0 0 1 ] = [ 2 0 0 0 2 0 0 0 2 ] 2I =2\begin{bmatrix} 1 &0 &0 \\ 0 & 1& 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 2 &0 &0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} 2 I = 2 ⎣ ⎡ 1 0 0 0 1 0 0 0 1 ⎦ ⎤ = ⎣ ⎡ 2 0 0 0 2 0 0 0 2 ⎦ ⎤
f ( A ) = [ − 1 0 0 0 − 2 1 0 − 2 − 3 ] + [ 2 0 0 0 2 0 0 0 2 ] = [ 1 0 0 0 0 1 0 − 2 − 1 ] = A f(A)=\begin{bmatrix} -1 &0 &0 \\ 0 & -2 & 1 \\ 0 & -2 & -3 \end{bmatrix} + \begin{bmatrix} 2 &0 &0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix} = \begin{bmatrix} 1 &0 &0 \\ 0 & 0& 1 \\ 0 & -2 & -1 \end{bmatrix} =A f ( A ) = ⎣ ⎡ − 1 0 0 0 − 2 − 2 0 1 − 3 ⎦ ⎤ + ⎣ ⎡ 2 0 0 0 2 0 0 0 2 ⎦ ⎤ = ⎣ ⎡ 1 0 0 0 0 − 2 0 1 − 1 ⎦ ⎤ = A
Therefore f ( T ) = T f(T)=T f ( T ) = T
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