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I've been trying to find material on matrix calculus but it seems hard to find ones with understandable proofs.

I'm doing research work and I am trying to verify some computation. Suppose that I have a matrix $A= \left( \begin{array}{ccc} \beta_{11} + c\beta_{12} +\beta_{13} & -c\beta_{12} & -\beta_{13} \\ -c\beta_{12} & c\beta_{12}+\beta_{22}+\beta_{23} & -\beta_{23} \\ -\beta_{13} & \beta_{23} & -\beta_{13}+\beta_{23}+\beta_{33} \end{array} \right) $ where $c$ is a constant, how do I evaluate $\frac{d}{d\beta_{12}}\det{(A)}$?

From what I have searched, if $A= \left( \begin{array}{ccc} \beta_{11} & \beta_{12} & \beta_{13} \\ \beta_{21} & \beta_{22} & \beta_{23} \\ \beta_{31} & \beta_{32} & \beta_{33} \end{array} \right) $ , i.e. no 2 elements are identical, then $\frac{d}{d\beta_{12}}\det{(A)}=\det{(A)}\cdot A^{-1}_{12}$.

What about the former case? Is there some sort of product rule like in 1-variable calculus?

I forgot to mention, the above is just a simplified case of the problem I'm working on. For my case, the matrix $A$ has dimension 300x300


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I mean you can straight up write out the determinant and take a partial's just a 3x3 matrix so this isn't terrible. – uncookedfalcon Dec 13 '12 at 8:39
Use Jacobi's formula. – copper.hat Dec 13 '12 at 8:42
hi guys thanks for the reply but I edited the question. The 3x3 matrix above is just a simplification to the actual problem I'm working on... – Tomas Jorovic Dec 13 '12 at 8:49

You can use the fact that $$\frac{\partial \det A}{\partial \alpha} = \det A \,\mathop{\rm tr} \left(A^{-1} \frac{\partial A}{\partial \alpha}\right), $$if $A$ is invertible.

Or more generally, you can use Jacobi's formula $$\frac{\partial \det A}{\partial \alpha}= \mathop{\rm tr} \left(\mathop{\rm adj}(A) \frac{\partial A}{\partial \alpha}\right).$$

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In your example, $\beta_{12}$ appears in a rank one fashion; row reductions may be done to find an equivalent formulation with only one appearance of the variable. If this is a common attribute to your general problem, that may be the way to go. In your example the first step would be the row reduction with respect to the variable $\beta_{12}$: $$\pmatrix{1 & 0 & 0 \\ 1&1&0 \\ 0&0&1}\left( \begin{array}{ccc} \beta_{11} + c\beta_{12} +\beta_{13} & -c\beta_{12} & -\beta_{13} \\ -c\beta_{12} & c\beta_{12}+\beta_{22}+\beta_{23} & -\beta_{23} \\ -\beta_{13} & \beta_{23} & -\beta_{13}+\beta_{23}+\beta_{33} \end{array} \right)=\left( \begin{array}{ccc} \beta_{11} + c\beta_{12} +\beta_{13} & -c\beta_{12} & -\beta_{13} \\ \beta_{11} + \beta_{13} & \beta_{22}+\beta_{23} & -\beta_{13} -\beta_{23} \\ -\beta_{13} & \beta_{23} & -\beta_{13}+\beta_{23}+\beta_{33} \end{array} \right)$$

$\beta_{12}$ now only appears in the first row. From here you could column reduce with respect to $\beta_{12}$ in order to use your formula regarding the single appearance of $\beta_{12}$.

You may also notice that when the variable appears in only one row that a similar formula regarding the dot product with the respective column of the inverse is possible.

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