10 I believe your proof is correct. Note that the best way of proving that $\det (A)=\det (A^t)$ depends very much on the definition of the determinant you are using. My personal favorite way of proving it is …

Typically we define determinants by a series of rules from which $\det (\alpha A)=\alpha^n\det (A)$ follows almost immediately. Even defining determinants as the expression used in Andrea's answer …

Aug 6, 2024 · 海运中，DEM和DET是两种常见的费用术语。DEM代表滞期费（Demurrage Charges），当船舶在港口停留超过预定时间，未能及时卸货或装货，船东会向租船人收取这部分费 …

Feb 7, 2020 · I was thinking about trying to argue because the numbers of a given matrix multiply as scalars, the determinant is the product of them all and because the order of the multiplication of det …

Sep 6, 2022 · Prove $$\\det \\left( e^A \\right) = e^{\\operatorname{tr}(A)}$$ for all matrices $A \\in \\mathbb{C}^{n \\times n}$.

Jul 12, 2016 · That actually makes sense to me now. Thank you for your assistance!

Aug 28, 2011 · Once you buy this interpretation of the determinant, $\det (AB)=\det (A)\det (B)$ follows immediately because the whole point of matrix multiplication is that $AB$ corresponds to the …

Jan 17, 2016 · Thus, its determinant will simply be the product of the diagonal entries, $ (\det A)^n$ Also, using the multiplicity of determinant function, we get $\det (A\cdot adjA) = \det A\cdot \det (adjA)$

Aug 28, 2018 · I see this in my textbook: I don't really understand the proof or why elementary matrices are involved. What is going on here?

Prove $\det (kA)=k^n\det A$ [closed] Ask Question Asked 12 years, 5 months ago Modified 6 years, 7 months ago