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  1. 线性代数 (det)是什么意思?_百度知道

    A矩阵的 行列式 (determinant),用符号det (A)表示。 行列式在数学中,是由解 线性方程组 产生的一种算式其 定义域 为nxn的矩阵 A,取值为一个 标量,写作det (A)或 | A | 。行列式可以看 …

  2. How do I prove that $\det A= \det A^T$? - Mathematics Stack …

    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 …

  3. 海运中DEM与DET分别代表什么?所谓的免堆与免用?所谓场内场 …

    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 …

  4. linear algebra - Why is $\det⁡ (-A)= (-1)^n\det (A)

    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 …

  5. prove that $\det (ABC) = \det (A) \det (B) \det (C)$ [for any $n×n ...

    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 …

  6. linear algebra - How to prove $\det \left (e^A\right) = e ...

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

  7. linear algebra - how does $\det ( (\det A) I)= (\det A)^n ...

    Apr 14, 2014 · $det (A)I$ is a diagonal matrix with all entries equal to $det (A)$. The determinant of a diagonal matrix can be found by multiplying the diagonal entries. So, we get the result.

  8. Why $\det (A^ {-1}) = 1/\det (A)$? - Mathematics Stack Exchange

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

  9. linear algebra - How to show that $\det (AB) =\det (A) \det (B ...

    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 …

  10. The determinant of adjugate matrix - Mathematics Stack Exchange

    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 …