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 …

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Prove $\det (kA)=k^n\det A$ [closed] Ask Question Asked 12 years, 4 months ago Modified 6 years, 7 months ago

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Feb 20, 2015 · It would be a good exercise to determine for which matrices, the identity $\det (A+b)=\det (A)+\det (B)$ holds. I think that it would work for rather few of them.

I know that there are three important results when taking the Determinants of Block matrices $$\\begin{align}\\det \\begin{bmatrix} A & B \\\\ 0 & D \\end ...

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 …

If $\text {det }\bf {A}=0$, this transformation is, in fact, a flattening (the geometric interpretation of the determinant is that it is the area produced by the transformation of the unit square): Any vector will …

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 …

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 $\det (A\cdot adjA) = \det A\cdot \det (adjA)$