Step-by-step Solution:

\(Statement A:\) If \( A \) and \( B \) are two invertible matrices of order \( n \), then their product is also invertible.
If \( A \) and \( B \) are invertible, their inverses exist.
The inverse of \( AB \) is given by: \[ (AB)^{-1} = B^{-1} A^{-1} \] Since \( B^{-1} A^{-1} \) also exists, \( AB \) is invertible.
Thus, Statement A is true.
\(Statement B:\) Every skew-symmetric matrix of odd order is invertible.
A skew-symmetric matrix \( A \) satisfies \( A^T = -A \).
The determinant of a skew-symmetric matrix of odd order is always zero, implying that it is singular (not invertible).
Thus, Statement B is false.
\(Statement C:\) If \( A \) is a non-singular matrix, then \( (A^T)^{-1} = (A^{-1})^T \).
The inverse of the transpose of a matrix satisfies: \[ (A^T)^{-1} = (A^{-1})^T \] This is a well-known property of matrix inverses.
Thus, Statement C is true.
\(Statement D:\) If \( A \) is an involutory matrix, then \( (I + A)(I - A) = 0 \).
A matrix \( A \) is involutory if: \[ A^2 = I \] Expanding the given expression: \[ (I + A)(I - A) = I - A + A - A^2 = I - A^2 \] Since \( A^2 = I \), we get: \[ I - I = 0 \] Thus, Statement D is true.
\(Statement E\): A diagonal matrix is both an upper triangular and a lower triangular matrix.
A diagonal matrix has nonzero elements only on its main diagonal.
Since all elements above and below the diagonal are zero, it is both upper triangular and lower triangular.
Thus, Statement E is true.
Final Answer:
True statements: A, C, D, E False statement: B Thus, the correct option is: \( {D} \quad \text{(A,C, D, E only)} \)