If A and B are invertible square matrices of order n, then which of the following is not correct?

For any two square matrices \(A\) and \(B\) of the same order, the determinant of their product is the product of their determinants, so \(\det(AB) = \det(A) \cdot \det(B)\).

If \(A\) is a square matrix of order \(n\) and \(k\) is a scalar, then multiplying \(A\) by \(k\) multiplies each of its \(n\) rows by \(k\). Thus, \(\det(kA) = k^n \det(A)\).

For an invertible matrix \(A\), \(A A^{-1} = I\). Taking the determinant on both sides gives \(\det(A) \cdot \det(A^{-1}) = \det(I) = 1\), which implies \(\det(A^{-1}) = \dfrac{1}{\det(A)}\).

However, the determinant of a sum of matrices is generally not equal to the sum of their determinants. Therefore, \(\det(A + B) = \det(A) + \det(B)\) is not correct.

Answer: \(\det(A + B) = \det(A) + \det(B)\)