Given that $A$ and $B$ are real invertible matrices such that $AB = -BA$, we want to find the correct relationship between their traces.

Analyzing the Property $AB = -BA$

1. Trace of a Matrix: The trace of a matrix $X$, denoted as $\text{Tr}(X)$, is the sum of its diagonal elements. A useful property of the trace function is that for any two matrices $X$ and $Y$, $\text{Tr}(XY) = \text{Tr}(YX).$

2. Applying the Trace to the Given Equation: Given $AB = -BA$, we can take the trace of both sides:

   $\text{Tr}(AB) = \text{Tr}(-BA).$

   Using the property of the trace, we have:

   $\text{Tr}(AB) = \text{Tr}(-BA) = -\text{Tr}(BA).$

   Since $\text{Tr}(AB) = \text{Tr}(BA)$, this gives:

   $\text{Tr}(AB) = -\text{Tr}(AB).$

   This implies:

   $\text{Tr}(AB) = 0.$

3. Conclusion on Traces of $A$ and $B$: Since $AB = -BA$ and both $A$ and $B$ are invertible, $A$ and $B$ must be matrices with specific properties. In particular, if they are real matrices and satisfy the anti-commutation relation $AB = -BA$, their traces must satisfy:

   $\text{Tr}(A) = 0 \quad \text{and} \quad \text{Tr}(B) = 0.$

   This is because the trace of a product of such matrices will always yield zero. Thus, both matrices have to be traceless.

Correct Answer:

The correct answer is:

c. $\text{Tr}(A) = \text{Tr}(B) = 0$.

Would you like more details or have any questions? Here are 5 related questions that could further enhance understanding:

1. Why is the trace of a product of matrices invariant under cyclic permutations?
2. Can you provide an example of two $2 \times 2$ matrices $A$ and $B$ such that $AB = -BA$ and both have a trace of zero?
3. What are the implications of a matrix being traceless in terms of its eigenvalues?
4. How does the property $AB = -BA$ influence the determinant of the matrices $A$ and $B$?
5. Are there any geometric interpretations for matrices that satisfy $AB = -BA$?

Tip: When dealing with matrix equations, always consider fundamental properties like trace, determinant, and rank for insight.