Using the iterative method to solve the system of equations $A X = b$, let $x^*$ be the exact solution, $\boldsymbol{X}^{(k)}$ the $k$th approximate solution, and $\varepsilon^{(k)} = X^{(k)} - X^{*}$, called the $k$th residual vector. Suppose
$$
A = \left[ \begin{array} {c r} 1 & -\frac{1}{2} \\ \\ -\frac{1}{2} & 1 \\ \end{array} \right], \quad b = \left[ \begin{array} {c r} 1 & 1 \\ -\frac{1}{2} & 1 \\ \end{array} \right]^{\mathrm{T}}.
$$
Try to the residual vector $\varepsilon^{(k)}$ for both the Jacobi iteration and the Gauss-Seidel iteration is().

Options:
A. J a c o b i:
$$
\varepsilon^{( k )}=
2^{-k} {\left[ \begin{array} {l l} {{1}} & {{0}} \\ {{0}} & {{1}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{2 \cdot2^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ {{{2^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ \end{array} \right]
$$
B. J a c o b i:
$$
\varepsilon^{( k )}=
7^{-k} {\left[ \begin{array} {l l} {{0}} & {{2}} \\ {{2}} & {{1}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{7 \cdot7^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ {{{7^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ \end{array} \right]
$$
C. J a c o b i:
$$
\varepsilon^{( k )}=
4^{-k} {\left[ \begin{array} {l l} {{0}} & {{2}} \\ {{2}} & {{0}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{4 \cdot4^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ {{{4^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ \end{array} \right]
$$
D. J a c o b i:
$$
\varepsilon^{( k )}=
3^{-k} {\left[ \begin{array} {l l} {{1}} & {{0}} \\ {{0}} & {{0}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{3 \cdot3^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ {{{3^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ \end{array} \right]
$$
E. J a c o b i:
$$
\varepsilon^{( k )}=
2^{-k} {\left[ \begin{array} {l l} {{1}} & {{1}} \\ {{0}} & {{1}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{2 \cdot2^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ {{{2^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ \end{array} \right]
$$
F. J a c o b i:
$$
\varepsilon^{( k )}=
4^{-k} {\left[ \begin{array} {l l} {{2}} & {{2}} \\ {{2}} & {{2}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{4 \cdot4^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ {{{4^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ \end{array} \right]
$$
G. J a c o b i:
$$
\varepsilon^{( k )}=
5^{-k} {\left[ \begin{array} {l l} {{0}} & {{1}} \\ {{1}} & {{2}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{5 \cdot5^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ {{{5^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ \end{array} \right]
$$
H. J a c o b i:
$$
\varepsilon^{( k )}=
6^{-k} {\left[ \begin{array} {l l} {{2}} & {{0}} \\ {{0}} & {{1}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{6 \cdot6^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ {{{6^{-2 k} \varepsilon{}_{1}^{( 0 )}}}} \\ \end{array} \right]
$$
I. J a c o b i:
$$
\varepsilon^{( k )}=
3^{-k} {\left[ \begin{array} {l l} {{1}} & {{1}} \\ {{1}} & {{1}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{3 \cdot3^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ {{{3^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ \end{array} \right]
$$
J. J a c o b i:
$$
\varepsilon^{( k )}=
2^{-k} {\left[ \begin{array} {l l} {{0}} & {{1}} \\ {{1}} & {{0}} \\ \end{array} \right]}^{k} {\varepsilon^{( 0 )}}
$$
G a u s s-S e i d e l:
$$
\varepsilon^{( k )}\mathcal{}=\left[ \begin{array} {c} {{{2 \cdot2^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ {{{2^{-2 k} \varepsilon{}_{2}^{( 0 )}}}} \\ \end{array} \right]
$$
Answer: J