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இங்கு <math> A_i </math> என்பது <math> A </math>யின் ''i''வது நிரலை நிரல் காவி <math> b </math> கொண்டு பிரதியிடுவதன் மூலம் உருவாகும் தாயமாகும்.
இவ்விதி மெய்யெண் புலம் மட்டுமன்றி எந்தவொரு புலத்திலும் குணகங்களையும் தெரியாக் கணியங்களையும் கொண்டுள்ள சமன்பாட்டுத் தொகுதிக்கும் பொருந்தும். It has recently been shown that Cramer's rule can be implemented in O(''n''<sup>3</sup>) time,<ref>Ken Habgood, Itamar Arel, A condensation-based application of Cramerʼs rule for solving large-scale linear systems, Journal of Discrete Algorithms, Volume 10, January 2012, Pages 98-109, ISSN 1570-8667, 10.1016/j.jda.2011.06.007.</ref> which is comparable to more common methods of solving systems of linear equations, such as [[Gaussian elimination]].
==Proof==
The proof for Cramer's rule uses just two properties of determinants: linearity with respect to any given column (taking for that column a [[linear combination]] of column vectors produces as determinant the corresponding linear combination of their determinants), and the fact that the determinant is zero whenever two columns are equal (the determinant is alternating in the columns).
Fix the index ''j'' of a column. Linearity means that if we consider only column ''j'' as variable (fixing the others arbitrarily), the resulting function {{math|'''R'''<sup>''n''</sup> → '''R'''}} (assuming matrix entries are in '''R''') can be given by a matrix, with one row and ''n'' columns. In fact this is precisely what [[Laplace expansion]] does, writing {{math|det(''A'') {{=}} ''C''<sub>1</sub>''a''<sub>1,''j''</sub> + … + ''C''<sub>''n''</sub>''a''<sub>''n'',''j''</sub>}} for certain coefficients ''C''<sub>1</sub>,…,''C''<sub>''n''</sub> that depend on the columns of ''A'' other than column ''j'' (the precise expression for these [[cofactor (linear algebra)|cofactor]]s is not important here). The value det(''A'') is then the result of applying the one-line matrix {{math|''L''<sub>(''j'')</sub> {{=}} (''C''<sub>1</sub> ''C''<sub>2</sub> … ''C''<sub>''n''</sub>)}} to column ''j'' of ''A''. If {{math|''L''<sub>(''j'')</sub>}} is applied to any ''other'' column ''k'' of ''A'', then the result is the determinant of the matrix obtained from ''A'' by replacing column ''j'' by a copy of column ''k'', which is 0 (the case of two equal columns).
Now consider a system of ''n'' linear equations in ''n'' unknowns <math>x_1, x_2,\ldots,x_n</math>, whose coefficient matrix is ''A'', with det(''A'') assumed to be nonzero:
:<math>\begin{matrix}a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n&=&b_1\\a_{21}x_1+a_{22}x_2+\cdots+a_{2n}x_n&=&b_2\\\vdots&\vdots&\vdots\\a_{n1}x_1+a_{n2}x_2+\cdots+a_{nn}x_n&=&b_n.\end{matrix}</math>
If one combines these equations by taking ''C''<sub>1</sub> times the first equation, plus ''C''<sub>2</sub> times the second, and so forth until ''C''<sub>''n''</sub> times the last, then the coefficient of ''x''<sub>''j''</sub> will become {{math|''C''<sub>1</sub>''a''<sub>1,''j''</sub> + … + ''C''<sub>''n''</sub>''a''<sub>''n'',''j''</sub> {{=}} det(''A'')}}, while the coefficients of all other unknowns become 0; the left hand side becomes simply det(''A'')''x''<sub>''j''</sub>. The right hand side is {{math|''C''<sub>1</sub>''b''<sub>1</sub> + … + ''C''<sub>''n''</sub>''b''<sub>''n''</sub>}}, which is {{math|''L''<sub>(''j'')</sub>}} applied to the column vector '''b''' of the right hand sides ''b''<sub>''i''</sub>. In fact what has been done here is multiply the matrix equation {{math|''A'' ⋅ '''x''' {{=}} '''b'''}} on the left by {{math|''L''<sub>(''j'')</sub>}}. Dividing by the nonzero number det(''A'') one finds the following equation, necessary to satisfy the system:
:<math>x_j=\frac{L_{(j)}\cdot\mathbf{b}}{\det(A)}.</math>
But by construction the numerator is determinant of the matrix obtained from ''A'' by replacing column ''j'' by '''b''', so we get the expression of Cramers rule as necessary condition for a solution. The same procedure can be repeated for other values of ''j'' to find values for the other unknowns.
The only point that remains to prove is that these values for the unknowns, the only possible ones, do indeed together form a solution. But if the matrix ''A'' is invertible with inverse ''A''<sup>−1</sup>, then {{math|'''x''' {{=}} ''A''<sup>−1</sup> ⋅ '''b'''}} will be a solution, thus showing its existence. To see that ''A'' is invertible when det(''A'') is nonzero, consider the ''n'' by ''n'' matrix ''M'' obtained by stacking the one-line matrices {{math|''L''<sub>(''j'')</sub>}} on top of each other for ''j'' = 1, 2, …, ''n'' (this gives the [[adjugate matrix]] for ''A''). It was shown that {{math|''L''<sub>(''j'')</sub> ⋅ ''A'' {{=}} (0 … 0 det(''A'') 0 … 0)}} where {{math|det(''A'')}} appears at the position ''j''; from this it follows that {{math|''M'' ⋅ ''A'' {{=}} det(''A'')''I''<sub>''n''</sub>}}. Therefore
:<math>\frac1{\det(A)}M=A^{-1},</math>
completing the proof.
==Finding inverse matrix==
{{main|Invertible matrix#Methods of matrix inversion}}
Let ''A'' be an ''n''×''n'' matrix. Then
:<math>\mathrm{Adj}(A)A = \mathrm{det}(A)I\,</math>
where Adj(''A'') denotes the [[adjugate matrix]] of ''A'', det(''A'') is the determinant, and ''I'' is the [[identity matrix]]. If det(''A'') is invertible in ''R'', then the inverse matrix of ''A'' is
:<math>A^{-1} = \frac{1}{\operatorname{det}(A)} \operatorname{Adj}(A).</math>
If ''R'' is a [[field (mathematics)|field]] (such as the field of real numbers), then this gives a formula for the inverse of ''A'', provided det(''A'') ≠ 0. In fact, this formula will work whenever ''R'' is a [[commutative ring]], provided that det(''A'') is a [[Unit (ring theory)|unit]]. If det(''A'') is not a unit, then ''A'' is not invertible.
==Applications==
===Explicit formulas for small systems===
Consider the linear system <math>\left\{\begin{matrix}ax+by&={\color{red}e}\\ cx + dy&= {\color{red}f}\end{matrix}\right.\ </math> which in matrix format is <math>\begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} {\color{red}e} \\ {\color{red}f} \end{bmatrix}.</math>
Assume ''ad-bc'' nonzero. Then, ''x'' and ''y'' can be found with Cramer's rule as
:<math>x = \begin{vmatrix} \color{red}{e} & b \\ \color{red}{f} & d \end{vmatrix}/\begin{vmatrix} a & b \\ c & d \end{vmatrix} = { {\color{red}e}d - b{\color{red}f} \over ad - bc}</math>
and
:<math>y = \begin{vmatrix} a & \color{red}{e} \\ c & \color{red}{f} \end{vmatrix}/\begin{vmatrix} a & b \\ c & d \end{vmatrix} = { a{\color{red}f} - {\color{red}e}c \over ad - bc}.</math>
The rules for 3×3 are similar. Given <math>\left\{\begin{matrix}ax + by + cz&= {\color{red}j}\\dx + ey + fz&= {\color{red}k}\\gx + hy + iz&= {\color{red}l}\end{matrix}\right.</math> which in matrix format is <math>\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} {\color{red}j} \\ {\color{red}k} \\ {\color{red}l} \end{bmatrix}.</math>
Then the values of ''x'', ''y'' and ''z'' can be found as follows:
:<math>x = \frac { \begin{vmatrix} {\color{red}j} & b & c \\ {\color{red}k} & e & f \\ {\color{red}l} & h & i \end{vmatrix} } { \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} }, \quad y = \frac { \begin{vmatrix} a & {\color{red}j} & c \\ d & {\color{red}k} & f \\ g & {\color{red}l} & i \end{vmatrix} } { \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} },\text{ and }z = \frac { \begin{vmatrix} a & b & {\color{red}j} \\ d & e & {\color{red}k} \\ g & h & {\color{red}l} \end{vmatrix} } { \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} }.</math>
===Differential geometry===
Cramer's rule is also extremely useful for solving problems in [[differential geometry]]. 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>
Finding an equation for <math>\dfrac{\partial x}{\partial u}</math> is a trivial application of Cramer's rule.
First, calculate the first derivatives of ''F'', ''G'', ''x'', and ''y'':
:<math>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</math>
:<math>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</math>
:<math>dx = \frac{\partial X}{\partial u} du + \frac{\partial X}{\partial v} dv</math>
:<math>dy = \frac{\partial Y}{\partial u} du + \frac{\partial Y}{\partial v} dv.</math>
Substituting ''dx'', ''dy'' into ''dF'' and ''dG'', we have:
:<math>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</math>
:<math>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.</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>\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}</math>
:<math>\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}</math>
:<math>\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}</math>
:<math>\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}.</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\left(F, G\right)}{\partial\left(u, y\right)}\right)}{\left(\frac{\partial\left(F, G\right)}{\partial\left(x, y\right)}\right)}.</math>
Similar formulae can be derived for <math>\frac{\partial x}{\partial v}</math>, <math>\frac{\partial y}{\partial u}</math>, <math>\frac{\partial y}{\partial v}.</math>
===Integer programming===
Cramer's rule can be used to prove that an [[integer programming]] problem whose constraint matrix is [[totally unimodular]] and whose right-hand side is integer, has integer basic solutions. This makes the integer program substantially easier to solve.
===Ordinary differential equations===
Cramer's rule is used to derive the general solution to an inhomogeneous linear differential equation by the method of [[variation of parameters]].
==Geometric interpretation==
[[File:Cramer.jpg|thumb|400px|Geometric interpretation of Cramer's rule. The areas of the second and third shaded parallelograms are the same and the second is <math>x_1</math> times the first. From this equality Cramer's rule follows.]]
Cramer's rule has a geometric interpretation that can be considered also a proof or simply giving insight about its geometric nature. These geometric arguments work in general and not only in the case of two equations with two unknowns presented here.
Given the system of equations
:<math>\begin{matrix}a_{11}x_1+a_{12}x_2&=b_1\\a_{21}x_1+a_{22}x_2&=b_2\end{matrix}</math>
it can be considered as an equation between vectors
:<math>x_1\binom{a_{11}}{a_{21}}+x_2\binom{a_{12}}{a_{22}}=\binom{b_1}{b_2}. </math>
The area of the parallelogram determined by <math>\binom{a_{11}}{a_{21}}</math> and <math>\binom{a_{12}}{a_{22}}</math> is given by the determinant of the system of equations:
:<math>\left|\begin{matrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{matrix}\right|.</math>
In general, when there are more variables and equations, the determinant of <math>n</math> vectors of length <math>n</math> will give the ''volume'' of the ''[[parallelepiped]]'' determined by those vectors in the <math>n</math>-th dimensional [[Euclidean space]].
Therefore the area of the parallelogram determined by <math>x_1\binom{a_{11}}{a_{21}}</math> and <math>\binom{a_{12}}{a_{22}}</math> has to be <math>x_1</math> times the area of the first one since one of the sides has been multiplied by this factor.
Now, this last parallelogram, by [[Cavalieri's principle]], has the same area as the parallelogram determined by <math>\binom{b_1}{b_2}=x_1\binom{a_{11}}{a_{21}}+x_2\binom{a_{12}}{a_{22}}</math> and <math>\binom{a_{12}}{a_{22}}</math>.
Equating the areas of this last and the second parallelogram gives the equation
:<math>\left|\begin{matrix}b_1&a_{12}\\b_2&a_{22}\end{matrix}\right|=\left|\begin{matrix}a_{11}x_1&a_{12}\\a_{21}x_1&a_{22}\end{matrix}\right|=x_1\left|\begin{matrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{matrix}\right|</math>
from which Cramer's rule follows.
==Incompatible and indeterminate cases==
A system of equations is said to be incompatible when there are no solutions and it is called indeterminate when there is more than one solution. For linear equations, an indeterminate system will have infinitely many solutions (if it is over an infinite field), since the solutions can be expressed in terms of one or more parameters that can take arbitrary values.
Cramer's rule applies to the case where the coefficient determinant is nonzero. In the contrary case the system is either incompatible or indeterminate, based on the values of the determinants only for 2x2 systems.
For 3x3 or higher systems, the only thing one can say when the coefficient determinant equals zero is: if any of the "numerator" determinants are nonzero, then the system must be incompatible. However, the converse is false: having all determinants zero does not imply that the system is indeterminate. A simple example where all determinants vanish but the system is still incompatible is the 3x3 system x+y+z=1, x+y+z=2, x+y+z=3.
==See also==
* [[Matrix (mathematics)|Matrix]]
==Notes==
{{Refimprove|date=August 2008}}
<references/>
==External links==
{{wikibooks|Linear Algebra/Cramer's Rule}}
* [http://planetmath.org/encyclopedia/ProofOfCramersRule.html Proof of Cramer's Rule]
* [http://sole.ooz.ie/ WebApp descriptively solving systems of linear equations with Cramer's Rule]
* [http://www.stud.feec.vutbr.cz/~xvapen02/vypocty/linrov.php?language=english Online Calculator of System of linear equations]
{{DEFAULTSORT:Cramer's Rule}}
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