https://byjus.com/jee/differentiation-integration-of-determinants/
I saw this and I can't understand how this formula was derived, like why can we integrate row wise and add up determinants? Is there an intuition/ proof behind this?
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See [this](https://math.stackexchange.com/q/668/721644). – Invisible May 24 '20 at 09:56
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what does that have to do with this – A plate of momos May 24 '20 at 11:01
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Hint: expanding $$ \det \left( \begin{matrix} f_1(x) & g_1(x) \\ f_2(x) & g_2(x) \end{matrix} \right) = f_1(x)g_2(x)-f_2(x)g_1(x) $$ and differentiating, $$f_1'(x)g_2(x)+f_1(x)g_2'(x)-f_2'(x)g_1(x)-f_2(x)g_1'(x),$$ you get the derivative of $\det \left( \begin{matrix} f_1(x) & g_1(x) \\ f_2(x) & g_2(x) \end{matrix} \right)$. You can rewrite the result as $$ \det \left( \begin{matrix} f_1'(x) & g_1'(x) \\ f_2(x) & g_2(x) \end{matrix} \right)+ \det \left( \begin{matrix} f_1(x) & g_1(x) \\ f_2'(x) & g_2'(x) \end{matrix} \right). $$ Use the same idea with integration.
Gibbs
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You should try yourself. Just apply the definition of determinant and integrate. – Gibbs May 24 '20 at 11:05
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By the Laplace expansion formula, a determinant can be expressed as a linear combination of the elements of a row (where the coefficients are the cofactors).
Hence by linearity of integration, you can integrate a row when the other rows are constants.
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If you could explain with an example, I think it would be better – A plate of momos May 24 '20 at 11:01