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This time I'm having trouble proving the following question:

Assuming that the elements of an $n\times n$-matrix $Y(t) = (y_{ij}(t))_{i,j=1,...,n}$ are differentiable, I need to show the following equality:

$$\frac{d}{dt}(\det(Y(t))) = \sum\limits_{1=1}^n \det\left(\begin{array}{ccc} y_{11} & \cdots & y_{1n}\\ \vdots & & \vdots \\ y_{i-1,1} & \cdots & y_{i-1,n}\\ y'_{i,1} & \cdots & y'_{i,n}\\ y_{i+1,1} & \cdots & y_{i+1,n}\\ \vdots & & \vdots \\ y_{n1} & \cdots & y_{nn}\\ \end{array}\right)$$

And as a hint I'm told that one way of proving this equality is to use the formula of determinant with n!, however I'm not sure I know this formula, and the only one I could think of was the one with Levi-Civita symbol:

$$\det(Y(t)) = \frac{1}{n!}\sum\limits_{i_1,...,i_n = 1; j_1,...,j_n = 1}\epsilon_{i_1,...,i_n}\epsilon_{j_1,...,j_n}\alpha_{i_1j_1,...,i_nj_n}$$

which I don't know how to derivate. Could anyone help me derivate this determinant ? any lue to start would be welcome, I didn't find anything related to my problem on mathstack.

Rhaena
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  • You must have made a mistake typing the right hand side of the identity you want to show: 1) the right hand side doesn't contain any derivatives of the functions $y_{ij}(t)$, which can't be right, and 2) the matrices just look like $Y$ itself with the first row replaced such that all entries are $y_{11}$. In the rows $i-1$, $i$ and $i+1$ you just replicated the original rows? – Christoph Apr 13 '21 at 16:36
  • I obviously did thanks for your attention ! I corrected those mistakes, and so to answer your last questions I only wrote the rows $i - 1$ and $i + 1$ because th $i$-th row is actually the one with derivatives – Rhaena Apr 13 '21 at 16:49
  • It helps @JeanMarie, but I think I have not seen enough of what is said to fully understand it at once, maybe after some carefull look I will find a solution thanks to that – Rhaena Apr 13 '21 at 17:16

2 Answers2

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You want to apply the product rule for $n$ factors: $$ \frac{\mathrm d}{\mathrm dt} (f_1(t)\, f_2(t) \cdots f_n(t)) = \sum_{i=1}^n f_1(t) \cdots f_{i-1}(t) \, f_i'(t) \, f_{i+1}(t) \cdots f_n(t). $$ Applying that to each summand of the Leibniz formula for $\det(Y(t))$ yields the desired result.

Christoph
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  • Thank you for your adivce, I have never used this product rule but I can see quite clearly how it applies for functions like polynomials or usual functions, however I'm not sur how to apply it for Leibniz formula since it contains both sum and product, so could you explain in a general way (so I can find myself how to apply it in my case after) how to use this product rule for a function defined with sum and product ? – Rhaena Apr 13 '21 at 17:14
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We have by definition \begin{align}\det(Y(t))&=\sum_{\sigma\in\mathfrak S_n}\varepsilon(\sigma)\prod_{i=1}^n y_{i,\sigma(i)}\\ \text{explicitly: }\hskip 4em &=\sum_{\sigma\in\mathfrak S_n}\varepsilon(\sigma)\,y_{1,\sigma(1)}y_{2,\sigma(2)}\dots y_{i,\sigma(i)}\dots y_{n,\sigma(n)}, \end{align} and differentiating with the product rule, we obtain \begin{align} \bigl(\det(Y(t))\bigr)'=\sum_{\sigma\in\mathfrak S_n}\varepsilon(\sigma)\Bigl[&y'_{1,\sigma(1)}y_{2,\sigma(2)}\dots y_{i,\sigma(i)}\dots y_{n,\sigma(n)}+y_{1,\sigma(1)}y'_{2,\sigma(2)}\dots y_{i,\sigma(i)}\dots y_{n,\sigma(n)}\\ &+\dots+y_{1,\sigma(1)}y_{2,\sigma(2)}\dots y'_{i,\sigma(i)}\dots y_{n,\sigma(n)}+\cdots\\ &+\dots+y_{1,\sigma(1)}y_{2,\sigma(2)}\dots y_{i,\sigma(i)}\dots y'_{n,\sigma(n)}\Bigr] \end{align}

Bernard
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