91Ó°ÊÓ

To find the dual code \(C^\perp\) of the given linear code C, we need to find all x in \(\mathbb{Z}_{2}^{n}\) such that \(x \cdot y = 0\) for all y in C. Given linear code C is represented by the matrix: $$\left(\begin{array}{ccccc} 1 & 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 & 1 \\ 1 & 0 & 0 & 1 & 0 \end{array}\right)$$ We can see that each row of C is a code word y. We need to find all x in \(\mathbb{Z}_{2}^{n}\) such that the dot product of x with each of these rows is 0. $$x = (x_{1},x_{2},x_{3},x_{4},x_{5})$$ We have the following equations: $$x_{1}+x_{2}+x_{3}=0$$ $$x_{3}+x_{5}=0$$ $$x_{1}+x_{4}=0$$

Solve the system of linear equations to find x values: $$x_{1}=x_{4}$$ $$x_{2}=-x_{1}-x_{3}$$ $$x_{3}=x_{5}$$ Now, set any arbitrary value to \(x_{1}\) and \(x_{3}\) (say, 0 or 1) and find corresponding values of \(x_{2}\), \(x_{4}\), and \(x_{5}\). After plugging these values back into the equations, we obtain four possible solutions for x: $$(0, 0, 0, 0, 0)$$ $$(1, 0, 0, 1, 0)$$ $$(0, 1, 0, 0, 1)$$ $$(1, 1, 0, 1, 1)$$ So, the dual code \(C^\perp\) is given by the following matrix: $$\left(\begin{array}{ccccc} 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \\ 1 & 1 & 0 & 1 & 1 \end{array}\right)$$

In order to show that the dual code \(C^\perp\) is an (n, n-k) -linear code, we will check that: 1. The sum of any two code words in \(C^\perp\) is also in \(C^\perp\). 2. The scalar multiple of any code words in \(C^\perp\) is also in \(C^\perp\). (1) We can see that elements in \(\mathbb{Z}_{2}\) are either 0 or 1, so the sum of any two elements of \(C^\perp\) in \(\mathbb{Z}_{2}\) is also an element in \(\mathbb{Z}_{2}\). Therefore, the sum of any two code words in \(C^\perp\) will be a code word in \(C^\perp\). (2) Since the elements of \(\mathbb{Z}_{2}\) are either 0 or 1, the scalar multiple of any of the code words in \(C^\perp\) will also be a code word in \(C^\perp\). As both of these conditions are satisfied, it can be concluded that \(C^\perp\) is an (n, n-k) -linear code.

Standard generator matrix G for C is: $$\left(\begin{array}{ccccc} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \end{array}\right)$$ Parity-check matrix H for C is: $$\left(\begin{array}{ccccc} 1 & 1 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 \end{array}\right)$$ Standard generator matrix G for \(C^\perp\) is: $$\left(\begin{array}{ccccc} 1 & 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 & 1 \end{array}\right)$$ Parity-check matrix H for \(C^\perp\) is: $$\left(\begin{array}{ccccc} 1 & 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 & 1 \\ 0 & 0 & 1 & 0 & 1 \end{array}\right)$$

In general, we can see that the standard generator matrix G of the code C becomes the parity-check matrix H of the dual code \(C^\perp\). On the other hand, the parity-check matrix H of the code C becomes the standard generator matrix G of the dual code \(C^\perp\). To prove this, let's assume the generator matrix G is of dimension (k, n) and H is of dimension (n-k, n). Since C is a (n, k)-linear code, we have: $$\textbf{G} \cdot \textbf{H}^T = \textbf{0}_{(k,n-k)}$$ Now the rows of the generator matrix G are code words of C, and their dot product with rows of H transpose is 0, which means rows of H are orthogonal to rows of G, i.e., rows of H are in \(C^\perp\). Since H has (n-k) rows, it forms a generator matrix for \(C^\perp\). Similarly, since rows of H are in \(C^\perp\), their dot product with rows of G will be 0, which means rows of G are orthogonal to rows of H, i.e., rows of G are in \(C^{\perp\perp}\). Since G has k rows, it forms a generator matrix for \(C^{\perp\perp}\). By the standard properties of the dual code, we know that \(C^{\perp\perp} = C\). Thus, we've shown that generator matrix G of the code C becomes the parity-check matrix H of the dual code \(C^\perp\), and parity-check matrix H of the code C becomes the generator matrix G of the dual code \(C^\perp\).