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If we have, \begin{equation} Z=\begin{bmatrix} \dfrac{\partial A}{\partial x} & \dfrac{\partial A}{\partial y}\\ \dfrac{\partial B}{\partial x} & \dfrac{\partial B}{\partial y} \end{bmatrix} \end{equation}

Where $ \dfrac{\partial A}{\partial x}, \dfrac{\partial A}{\partial y}, \dfrac{\partial B}{\partial x}$ and $\dfrac{\partial B}{\partial y}$ are matrices.

Now, in order to find $\dfrac{\partial x}{\partial A}$, should I invert $Z$ and get $\dfrac{\partial x}{\partial A}$ from the inverse (if yes please tell me how, I couldn't figure it out)? or I can simply inverse the submatrix $\dfrac{\partial x}{\partial A}=\left(\dfrac{\partial A}{\partial x}\right)^{-1}$?

Paul
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user594836
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  • A proper answer needs justification, so this is just a comment. You need the top left hand block of $Z^{-1}$, which equals $\dfrac{\partial B}{\partial y}/\det Z$. (I'm assuming the blocks are equal size.) – ancient mathematician Sep 25 '18 at 08:11
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    Are you trying to determine $(\partial A/ \partial x)^{-1} $ or something else? – Paul Sep 29 '18 at 10:11
  • @Paul He/She is trying to calculate $\dfrac{\partial x}{\partial A}$ (as I've understood the question) but seems to be confused whether to invert $Z$ matrix to get it or to invert $\dfrac{\partial A}{\partial x}$. – Hazem Sep 29 '18 at 11:28
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    @Hazem but what does $\partial x / \partial A$ mean? It looks like $A$ is supposed to be a matrix whose entries are functions of $x$, from the notation $\partial A / \partial x$. But then $\partial x / \partial A$ doesn't make sense. –  Sep 29 '18 at 12:06
  • @Brahadeesh You can consider $Z$ as a Jacobian matrix. I just found this answer but not sure if the answer applies in general or it is only applies for $ \rho =\sqrt{x^2+y^2}$, $\phi =\arctan2(y,x)$ or I can apply it everywhere. – user594836 Sep 30 '18 at 04:37
  • @Paul No, I want to calculate $\dfrac{\partial x}{\partial A}$ not $\dfrac{\partial A}{\partial x}^{-1}$ and still not sure if $\dfrac{\partial x}{\partial A}=\dfrac{\partial A}{\partial x}^{-1}$ though. – user594836 Sep 30 '18 at 04:41
  • In that case, you need to provide more context for the question to be clear. How did you arrive at the original matrix? Also, do you see that $\partial x / \partial A$ does not make sense when $A$ is a matrix? You need to have a definition for derivative of a variable w.r.t. a matrix, or change the question to one where $A$ and $B$ are functions, not matrices. This would also be in line with calling the original matrix a Jacobian matrix. –  Sep 30 '18 at 05:46
  • The answer you linked shows that the entries of the inverse of this matrix does not bear any relation (in general) to the inverses of the entries. –  Sep 30 '18 at 05:59
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    Are $x$ and $y$ two vectors? Are you trying to express $\frac{\partial x}{\partial A}$ as a matrix multiplication and summation of $\dfrac{\partial A}{\partial x}, \dfrac{\partial A}{\partial y}, \dfrac{\partial B}{\partial x}, \dfrac{\partial B}{\partial y}$ and $\dfrac{\partial y}{\partial B}$? – Hans Oct 01 '18 at 06:04
  • Is $A$ of the same dimension as $x$ and $B$ as $y$? In other words, are $\dfrac{\partial A}{\partial x}$ and $\dfrac{\partial B}{\partial y}$ square matrices? – Hans Oct 01 '18 at 20:24

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The question is not well posed. The OP has not specified whether $A$ is of the same dimension as $x$ and $B$ as $y$, or in other words, whether $\dfrac{\partial A}{\partial x}$ and $\dfrac{\partial B}{\partial y}$ are square matrices. Suppose the answer to the question is affirmative and in addition $\dfrac{\partial B}{\partial y}$ and $\dfrac{\partial A}{\partial x}$ are invertible. Let $a:=\dfrac{\partial A}{\partial x},\,b:=\dfrac{\partial A}{\partial y},\,c:=\dfrac{\partial B}{\partial x},\,d:=\dfrac{\partial B}{\partial y}$. By block Gaussian elimination we conclude that $\alpha:= a-bd^{-1}c$ and $\beta:= d-cd^{-1}b$ are invertible and $$Z^{-1}=\begin{bmatrix} \alpha^{-1} & -a^{-1}b\beta^{-1} \\ -d^{-1}c\alpha^{-1} & \beta^{-1} \end{bmatrix}.$$ We know $\dfrac{\partial x}{\partial A}=\alpha^{-1}$. The advantage of this algorithm is that all the inversions involved are for smaller size matrices and thus the complexity of the operation of which is greatly reduced especially for $Z$ of large dimension.

Hans
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