Jacobian of the inverse of a matrix times a vector

Question

I come across this in my research. If we know $\frac{\partial \mathbf{Ab_1}}{\partial \mathbf x} = \mathbf M$, where $A$ is a matrix that depends on $\mathbf x$, and $ \mathbf b_1$ is a vector that does not depend on $\mathbf x$. Is it possible to get $\frac{\partial \mathbf{A^{-1} b_2}}{\partial \mathbf x}$? $\mathbf b_2$ does not depend on $\mathbf x$ either.

Answer 1

No, it is not possible to get $\frac{\partial {A^{-1} b_2}}{\partial x}$ with the given hypothesis. Here is an example illustrating what can happen:

Consider $$ A = A(x) = \pmatrix{\text{e}^x & 0 \cr 0 & k\text{e}^x}, $$ where $k$ is a nonzero constant. Also let $b_1=\pmatrix{1\cr 0}$ and $b_2=\pmatrix{0\cr 1}$. We then have that $$ M = \frac{\partial {Ab_1}}{\partial x} = \frac{\partial}{\partial x} \pmatrix{\text{e}^x\cr 0} = \pmatrix{\text{e}^x\cr 0}. $$ On the other hand, the vector we are trying to "get", namely $$ \frac{\partial {A^{-1} b_2}}{\partial x} = \frac{\partial}{\partial x} \pmatrix{\text{e}^{-x} & 0 \cr 0 & \text{e}^{-x}/k} \pmatrix{0\cr 1} = \frac{\partial}{\partial x} \pmatrix{0\cr \text{e}^{-x}/k} = \pmatrix{0\cr -\text{e}^{-x}/k}, $$ involves knowledge of the value of $k$, which is not given a priory.