prove that $G=\left(\begin{array}{cc|cc} 1 & 0 &1 &1\\ 0 & 1 &1 &-1 \end{array}\right)$ is self dual

Question

I'm doing problem 6 in the book Fundamentals of Error correcting codes

the problem is to show that $G=\left(\begin{array}{cc|cc} 1 & 0 &1 &1\\ 0 & 1 &1 &-1 \end{array}\right)$ is self dual

I know that code is selfdual if $C=C^\perp$

And I know that $C^\perp=\{\mathbf{x}\in\mathbb{F_{q}^n}|\mathbf{x}\cdot \mathbf{c}=0 \forall \mathbf{c}\in C\}$

where $\mathbf{x}$ is if I'm not mistaken just $xG$ where x is a [1 k] matrix with information and G is the Generator matrix with size [k n]

So because C is supposed to be equal to $C^\perp$ then the $\mathbf{x}$ I choose must have the structure such that $\mathbf{x} \cdot \mathbf{x}=0$ and because these two vectors are [1 n] vectors this is the same as $\mathbf{x}\mathbf{x}^\mathbf{T}=0\Leftrightarrow xG(xG)^T=0\Leftrightarrow xGG^Tx^T=0$

Whish implyes that I'm done if $GG^T=0$

but I've calculated that $GG^T\neq 0$ So I assume that one or more of my arguments are wrong, but I can't find my own mistakes. Can you guys point them out?

Answer 1

We see that $GG^T=0$ if and only if $3=0$. So I strongly suspect that $p=3$, with $q=p^n$, and the ternary code is selfdual.