Find a generator polynomial and generator matrix of a triple error-correcting [15, 5] binary cyclic code with q=2, n=15, and k=5. You need to choose two cyclotomic cosets of size 4 and one cyclotomic coset of size 2 from the 2-cyclotomic cosets C0, C1, C3, C5, and C7. Since the code is triple error-correcting (d ≥ 7), the set T must contain δ – 1 ≥ 6 consecutive integers. For this specific code, T is chosen as T = C1 ∪ C3 ∪ C5 = {1, 2, 3, 4, 5, 6, 8, 9, 10, 12}.

Now, let α be a primitive element of IF2t where t = ord15(2) = 4. Consider the primitive binary polynomial x^4 + x + 1 of degree 4, i.e., α^4 + α + 1 = 0. The generator polynomial g(x) is constructed as the product of (x – αi) for each αi in T. Using the properties of α and αi, you can calculate M0(x), M1(x), M3(x), M5(x), and M7(x). Finally, the generator matrix of C is obtained as g(x) = M1(x)M3(x)M5(x) = (x^4 + x^3 + x^2 + x + 1)(x^2 + x + 1)(x^2 + x + 1).

This generator matrix defines the structure of the code and can be used for encoding and decoding in a [15, 5] binary cyclic code with triple error-correction capability.

The provided solution is correct for finding the generator polynomial and generator matrix of a triple error-correcting [15, 5] binary cyclic code with the specified parameters. The generator polynomial g(x) is correctly constructed as the product of (x – αi) for each αi in the chosen set T, and the generator matrix is obtained as the product of M1(x), M3(x), and M5(x). This generator matrix defines the structure of the code and can be used for encoding and decoding in this specific code.