Let $A$ and $B$ be two $n \times n$ matrices over real numbers. Let $\text{rank}(M)$ and $\det(M)$ denote the rank and determinant of a matrix $M$, respectively. Consider the following statements. I. $\text{rank}(AB) = \text{rank}(A) \text{rank}(B)$ II. $\det(AB) = \det(A) \det(B)$ III. $\text{rank}(A + B) \le \text{rank}(A) + \text{rank}(B)$ IV. $\det(A + B) \le \det(A) + \det(B)$ Which of the above statements are TRUE?

Let us analyze each statement for $n \times n$ real matrices $A$ and $B$:

• Statement I is FALSE: The rank of a product satisfies $\text{rank}(AB) \le \min(\text{rank}(A), \text{rank}(B))$. For example, if $A = B = I_2$, then $\text{rank}(AB) = 2$, but $\text{rank}(A)\text{rank}(B) = 2 \times 2 = 4$.

• Statement II is TRUE: By the multiplicative property of determinants, $\det(AB) = \det(A) \det(B)$ holds for all square matrices of the same dimension.

• Statement III is TRUE: By the subadditivity of matrix rank (since the column space of $A+B$ is contained in the sum of the column spaces of $A$ and $B$), $\text{rank}(A + B) \le \text{rank}(A) + \text{rank}(B)$.

• Statement IV is FALSE: The determinant is not generally subadditive. For instance, let $A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}$ and $B = \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$. Here $\det(A) = 0$ and $\det(B) = 0$, but $A+B = I_2$, giving $\det(A+B) = 1 > 0 + 0 = 0$.

Thus, statements II and III only are true, corresponding to option C.