La regla de Cramer es un teorema del álgebra lineal que da la solución de un sistema lineal de ecuaciones en términos de determinantes . Recibe este nombre en honor a Gabriel Cramer (1704-1752), quien publica la regla en su Introduction à l'analyse des lignes courbes algébriques de 1750, aunque Colin Maclaurin también publicó el método en su Treatise of Geometry de 1748 (y probablemente sabía del método desde 1729).[ 1] [ 2]
La regla de Cramer es de importancia teórica porque da una expresión explícita para la solución del sistema. Sin embargo, para sistemas de ecuaciones lineales de más de tres ecuaciones su aplicación para la resolución del mismo resulta excesivamente costosa: computacionalmente, es ineficiente para grandes matrices y por ello no es usado en aplicaciones prácticas que pueden implicar muchas ecuaciones. Sin embargo, como no es necesario pivotar matrices, es más eficiente que la eliminación gaussiana para matrices pequeñas, particularmente cuando son usadas operaciones SIMD .
Si
A
x
=
b
{\displaystyle \mathbf {Ax} =\mathbf {b} }
A
{\displaystyle \mathbf {A} }
x
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x
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,
…
,
x
n
)
{\displaystyle \mathbf {x} =(x_{1},\dots ,x_{n})}
vector columna de las incógnitas, y
b
{\displaystyle \mathbf {b} }
x
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=
det
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A
j
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det
(
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)
{\displaystyle x_{j}={\cfrac {\det(\mathbf {A} _{j})}{\det(\mathbf {A} )}}}
Donde
A
j
{\displaystyle \mathbf {A} _{j}}
j -ésima columna de
A
{\displaystyle \mathbf {A} }
b
{\displaystyle \mathbf {b} }
A
{\displaystyle \mathbf {A} }
Sea el siguiente sistema de dos ecuaciones lineales con dos incógnitas:
{
a
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+
b
y
=
e
c
x
+
d
y
=
f
{\displaystyle {\begin{cases}a{\color {blue}x}+b{\color {blue}y}={\color {red}e}\\c{\color {blue}x}+d{\color {blue}y}={\color {red}f}\end{cases}}}
Su representación matricial es
[
a
b
c
d
]
[
x
y
]
=
[
e
f
]
{\displaystyle {\begin{bmatrix}a&b\\c&d\end{bmatrix}}{\begin{bmatrix}{\color {blue}x}\\{\color {blue}y}\end{bmatrix}}={\begin{bmatrix}{\color {red}e}\\{\color {red}f}\end{bmatrix}}}
Si el sistema es compatible determinado, la solución viene dada, por la regla de Cramer, por los siguientes cocientes de determinantes :
x
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e
b
f
d
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=
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=
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{\displaystyle {\color {blue}x}={\frac {\begin{vmatrix}\color {red}{e}&b\\\color {red}{f}&d\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}}={\frac {{\color {red}e}d-b{\color {red}f}}{ad-bc}}\;,\quad {\color {blue}y}={\frac {\begin{vmatrix}a&\color {red}{e}\\c&\color {red}{f}\end{vmatrix}}{\begin{vmatrix}a&b\\c&d\end{vmatrix}}}={\frac {a{\color {red}f}-{\color {red}e}c}{ad-bc}}}
Ejemplo de la resolución de un sistema de dimensión 2x2:
Dado el sistema
{
3
x
+
y
=
9
2
x
+
3
y
=
13
{\displaystyle {\begin{cases}3x+y=9\\2x+3y=13\end{cases}}}
Su forma matricial es
[
3
1
2
3
]
[
x
y
]
=
[
9
13
]
{\displaystyle {\begin{bmatrix}3&1\\2&3\end{bmatrix}}{\begin{bmatrix}x\\y\end{bmatrix}}={\begin{bmatrix}9\\13\end{bmatrix}}}
Como el sistema es compatible determinado, por la regla de Cramer,
x
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9
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13
3
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=
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{\displaystyle x={\frac {\begin{vmatrix}9&1\\13&3\end{vmatrix}}{\begin{vmatrix}3&1\\2&3\end{vmatrix}}}={9\cdot 3-1\cdot 13 \over 3\cdot 3-1\cdot 2}=2}
y
=
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3
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2
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3
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3
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=
3
⋅
13
−
9
⋅
2
3
⋅
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−
1
⋅
2
=
3
{\displaystyle y={\frac {\begin{vmatrix}3&9\\2&13\end{vmatrix}}{\begin{vmatrix}3&1\\2&3\end{vmatrix}}}={3\cdot 13-9\cdot 2 \over 3\cdot 3-1\cdot 2}=3}
La regla para un sistema de 3x3, con una división de determinantes :
{
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+
b
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+
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z
=
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d
x
+
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g
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+
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+
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=
l
{\displaystyle {\begin{cases}a{\color {blue}x}+b{\color {blue}y}+c{\color {blue}z}={\color {red}j}\\d{\color {blue}x}+e{\color {blue}y}+f{\color {blue}z}={\color {red}k}\\g{\color {blue}x}+h{\color {blue}y}+i{\color {blue}z}={\color {red}l}\end{cases}}}
Que representadas en forma de matriz es:
[
a
b
c
d
e
f
g
h
i
]
[
x
y
z
]
=
[
j
k
l
]
{\displaystyle {\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}}{\begin{bmatrix}{\color {blue}x}\\{\color {blue}y}\\{\color {blue}z}\end{bmatrix}}={\begin{bmatrix}{\color {red}j}\\{\color {red}k}\\{\color {red}l}\end{bmatrix}}}
x
{\displaystyle x}
y
{\displaystyle y}
z
{\displaystyle z}
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{\displaystyle {\color {blue}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 {\color {blue}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}}}\;,\quad {\color {blue}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}}}}
Dado el sistema de ecuaciones lineales :
{
3
x
+
2
y
+
1
z
=
1
2
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+
0
y
+
1
z
=
2
−
1
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+
1
y
+
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z
=
4
{\displaystyle \left\lbrace \!\!\!{\begin{array}{rl}3x+2y+1z=&\!\!\!1\\2x+0y+1z=&\!\!\!2\\-1x+1y+2z=&\!\!\!4\end{array}}\right.}
Expresado en forma matricial :
[
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[
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[
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{\displaystyle {\begin{bmatrix}\,\,\,\,3&2&1\\\,\,\,\,2&0&1\\-1&1&2\end{bmatrix}}{\begin{bmatrix}x\\y\\z\end{bmatrix}}={\begin{bmatrix}1\\2\\4\end{bmatrix}}}
Los valores de
x
{\displaystyle x}
y
{\displaystyle y}
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{\displaystyle z}
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1
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+
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−
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11
;
{\displaystyle x={\frac {\begin{vmatrix}1&2&1\\2&0&1\\4&1&2\end{vmatrix}}{\begin{vmatrix}\,\,\,\,3&2&1\\\,\,\,\,2&0&1\\-1&1&2\end{vmatrix}}}={1\cdot 0\cdot 2+2\cdot 1\cdot 1+4\cdot 2\cdot 1-1\cdot 0\cdot 4-1\cdot 1\cdot 1-2\cdot 2\cdot 2 \over 3\cdot 0\cdot 2+2\cdot 1\cdot 1+(-1)\cdot 2\cdot 1-1\cdot 0\cdot (-1)-1\cdot 1\cdot 3-2\cdot 2\cdot 2}=-{\frac {1}{11}};\quad }
y
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11
;
{\displaystyle y={\frac {\begin{vmatrix}\,\,\,\,3&1&1\\\,\,\,\,2&2&1\\-1&4&2\end{vmatrix}}{\begin{vmatrix}\,\,\,\,3&2&1\\\,\,\,\,2&0&1\\-1&1&2\end{vmatrix}}}={3\cdot 2\cdot 2+2\cdot 4\cdot 1+(-1)\cdot 1\cdot 1-1\cdot 2\cdot (-1)-1\cdot 4\cdot 3-2\cdot 1\cdot 2 \over 3\cdot 0\cdot 2+2\cdot 1\cdot 1+(-1)\cdot 2\cdot 1-1\cdot 0\cdot (-1)-1\cdot 1\cdot 3-2\cdot 2\cdot 2}=-{\frac {5}{11}};\quad }
z
=
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3
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1
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⋅
0
⋅
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+
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+
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−
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⋅
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−
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−
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−
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⋅
2
⋅
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3
⋅
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⋅
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+
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+
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−
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⋅
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⋅
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=
24
11
{\displaystyle z={\frac {\begin{vmatrix}\,\,\,\,3&2&1\\\,\,\,\,2&0&2\\-1&1&4\end{vmatrix}}{\begin{vmatrix}\,\,\,\,3&2&1\\\,\,\,\,2&0&1\\-1&1&2\end{vmatrix}}}={3\cdot 0\cdot 4+2\cdot 1\cdot 1+(-1)\cdot 2\cdot 2-1\cdot 0\cdot (-1)-2\cdot 1\cdot 3-4\cdot 2\cdot 2 \over 3\cdot 0\cdot 2+2\cdot 1\cdot 1+(-1)\cdot 2\cdot 1-1\cdot 0\cdot (-1)-1\cdot 1\cdot 3-2\cdot 2\cdot 2}={\frac {24}{11}}}
Sean:
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=
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⋮
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)
{\displaystyle {\boldsymbol {x}}={\begin{pmatrix}x_{1}\\\vdots \\x_{n}\end{pmatrix}}\quad {\boldsymbol {b}}={\begin{pmatrix}b_{1}\\\vdots \\b_{n}\end{pmatrix}}}
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{\displaystyle {\boldsymbol {A}}_{j}=\left[{\begin{array}{llllllll}a_{1,1}&\cdots &a_{1,j-1}&b_{1}&a_{1,j+1}&\cdots &a_{1,n}\\a_{2,1}&\cdots &a_{2,j-1}&b_{2}&a_{2,j+1}&\cdots &a_{2,n}\\\\\vdots &&&\ddots &&&\vdots \\\\a_{n-1,1}&\cdots &a_{n-1,j-1}&b_{n-1}&a_{n-1,j+1}&\cdots &a_{n-1,n}\\a_{n,1}&\cdots &a_{n,j-1}&b_{n}&a_{n,j+1}&\cdots &a_{n,n}\end{array}}\right]}
Usando las propiedades de la multiplicación de matrices :
A
x
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b
⇔
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−
1
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=
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−
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⇔
I
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=
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−
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⇔
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=
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−
1
b
{\displaystyle {\boldsymbol {A}}{\boldsymbol {x}}={\boldsymbol {b}}\Leftrightarrow {\boldsymbol {A}}^{-1}{\boldsymbol {A}}{\boldsymbol {x}}={\boldsymbol {A}}^{-1}{\boldsymbol {b}}\Leftrightarrow {\boldsymbol {Ix}}={\boldsymbol {A}}^{-1}{\boldsymbol {b}}\Leftrightarrow {\boldsymbol {x}}={\boldsymbol {A}}^{-1}{\boldsymbol {b}}}
Entonces:
x
=
A
−
1
b
=
(
Adj
A
)
t
|
A
|
b
{\displaystyle {\boldsymbol {x}}={\boldsymbol {A}}^{-1}{\boldsymbol {b}}={\frac {(\operatorname {Adj} {\boldsymbol {A}})^{t}}{\left|{\boldsymbol {A}}\right|}}\;{\boldsymbol {b}}}
x
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(
A
−
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)
j
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,
{\displaystyle {\boldsymbol {x}}_{j}={\boldsymbol {(}}A^{-1}{\boldsymbol {b}})_{j}=\sum _{i=1}^{n}{\frac {{\boldsymbol {A}}_{ij}}{\left|{\boldsymbol {A}}\right|}}b_{i}={\frac {\sum _{i=1}^{n}{\boldsymbol {A}}_{ij}b_{i}}{\left|{\boldsymbol {A}}\right|}}={\cfrac {\left|{\boldsymbol {A}}_{j}\right|}{\left|{\boldsymbol {A}}\right|}},}
donde
A
i
j
{\displaystyle {\boldsymbol {A}}_{ij}}
i
j
{\displaystyle ij}
determinante de
A
j
{\displaystyle {\boldsymbol {A}}_{j}}
j
{\displaystyle j}
![{\displaystyle {\boldsymbol {A}}_{j}=\left[{\begin{array}{llllllll}a_{1,1}&\cdots &a_{1,j-1}&b_{1}&a_{1,j+1}&\cdots &a_{1,n}\\a_{2,1}&\cdots &a_{2,j-1}&b_{2}&a_{2,j+1}&\cdots &a_{2,n}\\\\\vdots &&&\ddots &&&\vdots \\\\a_{n-1,1}&\cdots &a_{n-1,j-1}&b_{n-1}&a_{n-1,j+1}&\cdots &a_{n-1,n}\\a_{n,1}&\cdots &a_{n,j-1}&b_{n}&a_{n,j+1}&\cdots &a_{n,n}\end{array}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/28d37de7ec5ed3042127590754d7bf02a4c0d35f)
Datos: Q322666
Multimedia: Cramer's rule / Q322666
