Editing Cramer's rule (section)

===Differential geometry===

====Ricci calculus====
Cramer's rule is used in the [[Ricci calculus]] in various calculations involving the [[Christoffel symbols]] of the first and second kind.<ref>{{Cite book|title=The Absolute Differential Calculus (Calculus of Tensors)|last=Levi-Civita|first=Tullio|publisher=Dover|year=1926|isbn=9780486634012|pages=111–112}}</ref>

In particular, Cramer's rule can be used to prove that the divergence operator on a [[Riemannian manifold]] is invariant with respect to change of coordinates. We give a direct proof, suppressing the role of the Christoffel symbols.
Let <math>(M,g)</math> be a Riemannian manifold equipped with [[Manifold#Charts|local coordinates]] <math> (x^1, x^2, \dots, x^n)</math>. Let <math>A=A^i \frac{\partial}{\partial x^i}</math> be a [[vector field]]. We use the [[Einstein notation|summation convention]] throughout.

:'''Theorem'''.
:''The ''divergence'' of <math>A</math>,''
::<math> \operatorname{div} A = \frac{1}{\sqrt{\det g}} \frac{\partial}{\partial x^i} \left( A^i \sqrt{\det g} \right),</math>
:''is invariant under change of coordinates.''

{{Collapse top|title=''Proof''}}
Let <math>(x^1,x^2,\ldots,x^n)\mapsto (\bar x^1,\ldots,\bar x^n)</math> be a [[coordinate transformation]] with [[invertible matrix|non-singular]] [[Jacobian matrix and determinant|Jacobian]].  Then the classical [[Vector field#Coordinate transformation law|transformation laws]] imply that <math>A=\bar A^{k}\frac{\partial}{\partial\bar x^{k}}</math> where <math>\bar A^{k}=\frac{\partial \bar x^{k}}{\partial x^{j}}A^{j}</math>.  Similarly, if <math>g=g_{mk}\,dx^{m}\otimes dx^{k}=\bar{g}_{ij}\,d\bar x^{i}\otimes d\bar x^{j}</math>, then <math>\bar{g}_{ij}=\,\frac{\partial x^{m}}{\partial\bar x^{i}}\frac{\partial x^{k}}{\partial \bar x^{j}}g_{mk}</math>.  
Writing this transformation law in terms of matrices yields <math>\bar g=\left(\frac{\partial x}{\partial\bar{x}}\right)^{\text{T}}g\left(\frac{\partial x}{\partial\bar{x}}\right)</math>, which implies <math>\det\bar g=\left(\det\left(\frac{\partial x}{\partial\bar{x}}\right)\right)^{2}\det g</math>.

Now one computes
:<math>\begin{align}
\operatorname{div} A &=\frac{1}{\sqrt{\det g}}\frac{\partial}{\partial x^{i}}\left( A^{i}\sqrt{\det g}\right)\\
	&=\det\left(\frac{\partial x}{\partial\bar{x}}\right)\frac{1}{\sqrt{\det\bar g}}\frac{\partial \bar x^k}{\partial x^{i}}\frac{\partial}{\partial\bar x^{k}}\left(\frac{\partial x^{i}}{\partial \bar x^{\ell}}\bar{A}^{\ell}\det\!\left(\frac{\partial x}{\partial\bar{x}}\right)^{\!\!-1}\!\sqrt{\det\bar g}\right).
\end{align}</math>
In order to show that this equals 
<math>\frac{1}{\sqrt{\det\bar g}}\frac{\partial}{\partial\bar x^{k}}\left(\bar A^{k}\sqrt{\det\bar{g}}\right)</math>,
it is necessary and sufficient to show that 
:<math>\frac{\partial\bar x^{k}}{\partial x^{i}}\frac{\partial}{\partial\bar x^{k}}\left(\frac{\partial x^{i}}{\partial \bar x^{\ell}}\det\!\left(\frac{\partial x}{\partial\bar{x}}\right)^{\!\!\!-1}\right)=0\qquad\text{for all } \ell, </math>
which is equivalent to
:<math>\frac{\partial}{\partial \bar x^{\ell}}\det\left(\frac{\partial x}{\partial\bar{x}}\right)
=\det\left(\frac{\partial x}{\partial\bar{x}}\right)\frac{\partial\bar x^{k}}{\partial x^{i}}\frac{\partial^{2}x^{i}}{\partial\bar x^{k}\partial\bar x^{\ell}}.
</math>
Carrying out the differentiation on the left-hand side, we get:
:<math>\begin{align}
	\frac{\partial}{\partial\bar x^{\ell}}\det\left(\frac{\partial x}{\partial\bar{x}}\right)
	&=(-1)^{i+j}\frac{\partial^{2}x^{i}}{\partial\bar x^{\ell}\partial\bar x^{j}}\det M(i|j)\\
	&=\frac{\partial^{2}x^{i}}{\partial\bar x^{\ell}\partial\bar x^{j}}\det\left(\frac{\partial x}{\partial\bar{x}}\right)\frac{(-1)^{i+j}}{\det\left(\frac{\partial x}{\partial\bar{x}}\right)}\det M(i|j)=(\ast),
	\end{align}</math>
where <math>M(i|j)</math> denotes the matrix obtained from <math>\left(\frac{\partial x}{\partial\bar{x}}\right)</math> by deleting the <math>i</math>th row and <math>j</math>th column.
But Cramer's Rule says that 
:<math>\frac{(-1)^{i+j}}{\det\left(\frac{\partial x}{\partial\bar{x}}\right)}\det M(i|j) </math>
is the <math>(j,i)</math>th entry of the matrix <math>\left(\frac{\partial \bar{x}}{\partial x}\right)</math>.
Thus
:<math>(\ast)=\det\left(\frac{\partial x}{\partial\bar{x}}\right)\frac{\partial^{2}x^{i}}{\partial\bar x^{\ell}\partial\bar x^{j}}\frac{\partial\bar x^{j}}{\partial x^{i}},</math>
completing the proof.

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====Computing derivatives implicitly====
Consider the two equations <math>F(x, y, u, v) = 0</math> and <math>G(x, y, u, v) = 0</math>. When ''u'' and ''v'' are independent variables, we can define <math>x = X(u, v)</math> and <math>y = Y(u, v).</math>

An equation for <math>\dfrac{\partial x}{\partial u}</math> can be found by applying Cramer's rule.

{{Collapse top|title=''Calculation of <math>\dfrac{\partial x}{\partial u}</math>''}}
First, calculate the first derivatives of ''F'', ''G'', ''x'', and ''y'':

:<math>\begin{align}
dF &= \frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy +\frac{\partial F}{\partial u} du +\frac{\partial F}{\partial v} dv = 0 \\[6pt]
dG &= \frac{\partial G}{\partial x} dx + \frac{\partial G}{\partial y} dy +\frac{\partial G}{\partial u} du +\frac{\partial G}{\partial v} dv = 0 \\[6pt]
dx &= \frac{\partial X}{\partial u} du + \frac{\partial X}{\partial v} dv \\[6pt]
dy &= \frac{\partial Y}{\partial u} du + \frac{\partial Y}{\partial v} dv.
\end{align}</math>

Substituting ''dx'', ''dy'' into ''dF'' and ''dG'', we have:

:<math>\begin{align}
dF &= \left(\frac{\partial F}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial u} + \frac{\partial F}{\partial u} \right) du + \left(\frac{\partial F}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial v} +\frac{\partial F}{\partial v} \right) dv = 0 \\ [6pt]
dG &= \left(\frac{\partial G}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial u} +\frac{\partial G}{\partial u} \right) du + \left(\frac{\partial G}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial v} +\frac{\partial G}{\partial v} \right) dv = 0.
\end{align}</math>

Since ''u'', ''v'' are both independent, the coefficients of ''du'', ''dv'' must be zero.  So we can write out equations for the coefficients:

:<math>\begin{align}
\frac{\partial F}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial u} & = -\frac{\partial F}{\partial u} \\[6pt]
\frac{\partial G}{\partial x} \frac{\partial x}{\partial u} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial u} & = -\frac{\partial G}{\partial u} \\[6pt]
\frac{\partial F}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial F}{\partial y} \frac{\partial y}{\partial v} & = -\frac{\partial F}{\partial v} \\[6pt]
\frac{\partial G}{\partial x} \frac{\partial x}{\partial v} +\frac{\partial G}{\partial y} \frac{\partial y}{\partial v} & = -\frac{\partial G}{\partial v}.
\end{align}</math>

Now, by Cramer's rule, we see that:

:<math>\frac{\partial x}{\partial u} = \frac{\begin{vmatrix} -\frac{\partial F}{\partial u} & \frac{\partial F}{\partial y} \\ -\frac{\partial G}{\partial u} & \frac{\partial G}{\partial y}\end{vmatrix}}{\begin{vmatrix}\frac{\partial F}{\partial x} & \frac{\partial F}{\partial y} \\ \frac{\partial G}{\partial x} & \frac{\partial G}{\partial y}\end{vmatrix}}.</math>

This is now a formula in terms of two [[Jacobian matrix and determinant|Jacobian]]s:

:<math>\frac{\partial x}{\partial u} = -\frac{\left(\frac{\partial (F, G)}{\partial (u, y)}\right)}{\left(\frac{\partial (F, G)}{\partial(x, y)}\right)}.</math>

Similar formulas can be derived for <math>\frac{\partial x}{\partial v}, \frac{\partial y}{\partial u}, \frac{\partial y}{\partial v}.</math>
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