# Determinant of the sum of matrices: $\det (A + B^T) = \det(A^T + B)$ [closed]

How can you show that $$\det(A + B^T) = \det(A^T + B)$$ for any $n\times n$ matrices $A$ and $B.$

-

## closed as off-topic by Najib Idrissi, SchrodingersCat, Davide Giraudo, Harish Chandra Rajpoot, Jack's wasted lifeFeb 9 at 16:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Najib Idrissi, SchrodingersCat, Davide Giraudo, Harish Chandra Rajpoot, Jack's wasted life
If this question can be reworded to fit the rules in the help center, please edit the question.

Prove that $$(A+B)^T=A^T+B^T$$ and that

$$\det A^T=\det A$$

and you're done. For the second one, consider expanding the determinant "column-wise" in one case, and "row wise" in the other. In fact, you can use the cofactor expansion of the determinant to prove the claim by induction.

-

Hint: we have $(A^T+B)^T=A+B^T$

-

$$\det(A)=\det(A^T)$$

$$\det(A + B^T) = \det((A+B^T)^T) = \det(A^T +B)$$

-