Let $M$ be an invertible $n \times n$ real matrix. Which of the following statements are necessarily TRUE for any such matrix $M$?
Options:
(A) $\det(M^{-1}) = \frac{1}{\det(M)}$
(B) $\det(kM) = k^n \det(M)$ for any real scalar $k$
(C) $\det(M^T M) = (\det M)^2 > 0$
(D) The trace of $M$ is always strictly positive
Answer Key:
A, B, C
Step-by-Step Solution & Explanation
1. For an invertible matrix $M$, $M M^{-1} = I \implies \det(M)\det(M^{-1}) = \det(I) = 1 \implies \det(M^{-1}) = 1/\det(M)$. (Statement A is True)
2. By properties of determinants of $n \times n$ matrices, factoring scalar $k$ from each of the $n$ rows yields $\det(kM) = k^n \det(M)$. (Statement B is True)
3. $\det(M^T M) = \det(M^T)\det(M) = (\det M)(\det M) = (\det M)^2$. Since $M$ is invertible, $\det(M) \neq 0$, so $(\det M)^2 > 0$. (Statement C is True)
4. The trace can be negative or zero (e.g. $M = -I_n$ has $\text{Tr}(M) = -n < 0$). Hence Statement D is False.