$\therefore$ $H = \left [\begin{array} 11 0 0 \\ 0 1 1 \\ 1 1 1 \\ 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array} \right] $

be a parity check, matrix, Determine the group code $e_H : B^3 \rightarrow B^6$

Solution: Let,

Here, n = 6, r = 3 $\rightarrow$ m = n - r = 3

$\therefore$ Encoding function $e : B^3 \rightarrow B^6$

e(000) = 000 $x_1 x_2 x_3$

e(001) = 001 $x_1 x_2 x_3$

e(010) = 010 $x_1 x_2 x_3$

e(011) = 011 $x_1 x_2 x_3$

e(100) = 100 $x_1 x_2 x_3$

e(101) = 101 $x_1 x_2 x_3$

e(110) = 110 $x_1 x_2 x_3$

e(111) = 111 $x_1 x_2 x_3$

1] e(000) = [000] $\begin{matrix} \Bigg [1 0 0\\ 0 1 1\\ 1 1 1\Bigg] \end{matrix}$

= 000

$\therefore$ e = (000) = 000 000

2] e(001) = [001] $\begin{matrix} \Bigg [1 0 0\\ 0 1 1\\ 1 1 1\Bigg] \end{matrix}$

= 111

e (001) = 001111

3] e (010) = [010] $\begin{matrix} \Bigg [1 0 0\\ 0 1 1\\ 1 1 1\Bigg] \end{matrix}$

= 011

e(010) = 010011

4] e(011) = [011] $\begin{matrix} \Bigg [1 0 0\\ 0 1 1\\ 1 1 1\Bigg] \end{matrix}$

= 100

e(011) = 011100

5] e(100) = [100] $\begin{matrix} \Bigg [1 0 0\\ 0 1 1\\ 1 1 1\Bigg] \end{matrix}$

= 011

e(100) = 100100

6] e(101) = [101] $\begin{matrix} \Bigg [1 0 0\\ 0 1 1\\ 1 1 1\Bigg] \end{matrix}$

= 011

e(101) = 101011

7] e(110) = [110] $\begin{matrix} \Bigg [1 0 0\\ 0 1 1\\ 1 1 1\Bigg] \end{matrix}$

= 111

e(110) = 110111

8] e(111) = [111] $\begin{matrix} \Bigg [1 0 0\\ 0 1 1\\ 1 1 1\Bigg] \end{matrix}$

= 000

e(111) = 111000

$\therefore$ Group code $e_n : B^3 \rightarrow B^5$ is

e(000) = 000000

e(001) = 001111

e(010) = 010011

e(011) = 011100

e(100) = 100100

e(101) = 101011

e(110) = 110111

e(111) = 111000