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

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

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## put on hold as off-topic by Najib Idrissi, SchrodingersCat, Davide Giraudo, Harish Chandra Rajpoot, Jack's wasted life12 hours ago

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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.

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

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

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

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