How do I prove that $\det A= \det A^T$? - Mathematics Stack Exchange 9 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 by giving a definition of the determinant such that $\det (A)=\det (A^t)$ is obviously true
prove that $\det (ABC) = \det (A) \det (B) \det (C)$ [for any $n×n . . . 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
$\det (I+A) = 1 + tr (A) + \det (A)$ for $n=2$ and for $n gt;2$? You'll need to complete a few actions and gain 15 reputation points before being able to upvote Upvoting indicates when questions and answers are useful What's reputation and how do I get it? Instead, you can save this post to reference later
linear algebra - How to show that $\det (AB) =\det (A) \det (B . . . 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 $A \circ B$