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Given that $(uA+C)\mathbf{x}=\mathbf{b}$ where only $u\in \mathbb{R}$ and $\mathbf{x}\in\mathbb{R}^n$ are unknowns, and where $(uA+C)\in\mathbb{R}^{n\times n}$ is an invertible matrix, how can I determine $\frac{d\mathbf{x}}{du}$?

I rewrite the equation to $$\mathbf{x}=(uA+C)^{-1}\mathbf{b}$$

and wonder whether there is any way to find/simplify

$$\frac{d}{du}(uA+C)^{-1}\mathbf{b}$$

Background

In my particular case $(uA+C)\mathbf{x} = \mathbf{b}$ comes from

$$ \begin{bmatrix} -x_1 & -y_1 & -1 & 0 & 0 & 0 & x_1x_1' & y_1x_1' & x_1' \\ 0 & 0 & 0 & -x_1 & -y_1 & -1 & x_1y_1' & y_1y_1' & y_1' \\ -x_2 & -y_2 & -1 & 0 & 0 & 0 & x_2x_2' & y_2x_2' & x_2' \\ 0 & 0 & 0 & -x_2 & -y_2 & -1 & x_2y_2' & y_2y_2' & y_2' \\ -x_3 & -y_3 & -1 & 0 & 0 & 0 & x_3x_3' & y_3x_3' & x_3' \\ 0 & 0 & 0 & -x_3 & -y_3 & -1 & x_3y_3' & y_3y_3' & y_3' \\ -x_4 & -y_4 & -1 & 0 & 0 & 0 & x_4x_4' & y_4x_4' & x_4' \\ 0 & 0 & 0 & -x_4 & -y_4 & -1 & x_4y_4' & y_4y_4' & y_4' \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix}h1 \\ h2 \\ h3 \\ h4 \\ h5 \\ h6 \\ h7 \\ h8 \\h9 \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\1 \end{bmatrix}$$

(which comes from here)

where $u$ is one of $x_1'$, $y_1'$, $x_2'$, $y_2'$, $x_3'$, $y_3'$, $x_4'$, $y_4'$. For example, for $u\equiv x_1'$ we have

$$ A = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & x_1 & y_1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix} $$

Edit

I vaguely remember a technique called implicit differentiation which I feel may be useful:

$$ \frac{d}{du}(uA+C)\mathbf{x}=\frac{d}{du}\mathbf{b} $$ $$ \frac{d}{du}uA\mathbf{x}+\frac{d}{du}C\mathbf{x}=\mathbf{0} $$ $$ A\frac{d}{du}u\mathbf{x}+C\frac{d\mathbf{x}}{du}=\mathbf{0} $$ $$ A(\mathbf{x}+u\frac{d\mathbf{x}}{du})+C\frac{d\mathbf{x}}{du}=\mathbf{0} $$ $$ A\mathbf{x}+(uA+C)\frac{d\mathbf{x}}{du}=\mathbf{0} $$ $$ \frac{d\mathbf{x}}{du}=-(uA+C)^{-1}A\mathbf{x} $$

... did I just solve it; is this correct?

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  • $\begingroup$ Yes, you just solved it! $\endgroup$ – Robert Lewis Apr 26 '20 at 0:20
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Define $\,M=(C+uA)\,$ then the given equation becomes $\,Mx=b$

Differentiate the equation (with respect to $u)\,$ then solve for $\dot x=\left(\frac{dx}{du}\right)$ $$\eqalign{ \dot Mx + M\dot x = \dot b \\ Ax + M\dot x = 0 \\ \dot x = -M^{-1}Ax \\ }$$ This is indeed the implicit differentiation technique that you remembered.

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Hint Since $b$ does not depend on $u$, $$\frac{d}{du}[(u A + C)^{-1} {\bf b}] = \frac{d}{du}[(u A + C)^{-1}] {\bf b} ,$$ and so it suffices to know how to compute the derivative $\frac{d}{du} [P(u)^{-1}]$ inverse of a matrix function $$P : \Bbb R \to M_n (\Bbb R)$$ (wherever that inverse is defined).

We we can find $\frac{d}{du}(P(u)^{-1})$ in terms of $P$ and $\frac{d P}{dt}$ by differentiating both sides of $P(u) P(u)^{-1} = I$ and isolating $\frac{d}{du}[P(u)^{-1}]$.

Suppressing the argument $u$, we have $$\frac{dP}{du} P^{-1} + P \frac{d}{du} (P^{-1}) , $$ so $$\frac{d}{du} (P^{-1}) = - P^{-1} \frac{dP}{du} P^{-1} .$$

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Start with the given equation

$(uA + C)\mathbf x = \mathbf b; \tag 1$

since $(uA + C)$ is invertible we may directly write

$\mathbf x = (uA + C)^{-1} \mathbf b \tag 2$

which expresses $\mathbf x$ as a function of $u$; then

$\mathbf x' = ((uA + C)^{-1})' \mathbf b; \tag 3$

we may compute $((uA + C)^{-1})'$ as follows: for any parametrized invertible matrix $Y(u)$ we write

$YY^{-1} = I, \tag 4$

and differentiate:

$Y'Y^{-1} + Y(Y^{-1})'= 0, \tag 5$

or

$Y'Y^{-1} = -Y(Y^{-1})', \tag 6$

from which we immediately obtain

$(Y^{-1})' = -Y^{-1}Y'Y^{-1}; \tag 7$

taking

$Y(u) = uA + C \tag 8$

we arrive at

$((uA + C)^{-1})' = (uA + C)^{-1}A(uA + C)^{-1}, \tag 9$

whence from (3)

$\mathbf x' = (uA + C)^{-1} A (uA + C)^{-1} \mathbf b, \tag{10}$

and in light of (2),

$\mathbf x' = (uA + C)^{-1} A \mathbf x. \tag{11}$

There is in fact a much shorter route to this result if one accepts that $\mathbf x(u)$ is differentiable, as has indeed been proved in the above; in that event we may simply differentiate (1) and find

$(uA + C)'\mathbf x + (uA + C)\mathbf x' = 0, \tag{12}$

whence we directly write

$\mathbf x' = -(uA + C)^{-1}A \mathbf x, \tag{13}$

and via (1),

$\mathbf x' = -(uA + C)^{-1}A (uA + C)^{-1} \mathbf b. \tag{14}$

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