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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 | 。行列式可以看做是有向面积或体积。 扩展资料 性质 ①行列式A中某行 (或列)用同一数k乘,其结果等于kA。 ②行列式A等于其转置. With this, you can obtain formula similar to what you have for $n = 2$: 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用。而DET则是拘留费（Detention Charge），当集装箱在目的港滞留超过免堆期（如Free Detention），即使货物尚未清关，也可能产生额外费用.
10 i believe your proof is correct When $a$ is a $n \times n$ matrix, the above expansion terminate at the $t^n$ term with coefficient equal to $\det a$ 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 by giving a definition of the determinant such that $\det (a)=\det (a^t)$ is obviously true.
Thus, its determinant will simply be the product of the diagonal entries, $ (\det a)^n$ also, using the multiplicity of determinant function, we get $\det (a\cdot adja) = \det a\cdot \det (adja)$ 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 gives this right away. 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 (abc) stays the same, det (abc) = det (a) det (b) det (c) holds true
However, i don't think is a good enough proof and would greatly appreciate some insight. DET（Detention Charge）：拘留费 2.免堆期（FREE DETENTION），意思是免去集装箱在目的港码头的堆存费用，主要是发生在收货人不能迅速清关，或因为货量较大的情况下，需要在目的港申请减免此费用。 Prove $$\\det \\left( e^a \\right) = e^{\\operatorname{tr}(a)}$$ for all matrices $a \\in \\mathbb{c}^{n \\times n}$. 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 composed linear transformation $a \circ b$.
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