Mějme soustavu lineárních rovnic , která obsahuje stejně neznámých jako rovnic. Označme matici soustavy
A
{\displaystyle \mathbf {A} }
n
×
n
{\displaystyle n\times n}
A
i
{\displaystyle \mathbf {A} _{i}}
matici , kterou získáme z matice
A
{\displaystyle \mathbf {A} }
i
{\displaystyle i}
Pokud zapíšeme matice soustavy a vektor pravých stran jako
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{\displaystyle \mathbf {A} ={\begin{pmatrix}a_{11}&a_{12}&\cdots &a_{1n}\\a_{21}&a_{22}&\cdots &a_{2n}\\\vdots &\vdots &\ddots &\vdots \\a_{n1}&a_{n2}&\cdots &a_{nn}\end{pmatrix}}}
B
=
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b
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b
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⋮
b
n
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{\displaystyle \mathbf {B} ={\begin{pmatrix}b_{1}\\b_{2}\\\vdots \\b_{n}\end{pmatrix}}}
pak má
A
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{\displaystyle \mathbf {A} _{i}}
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{\displaystyle \mathbf {A} _{i}={\begin{pmatrix}a_{11}&a_{12}&\cdots &a_{1,i-1}&b_{1}&a_{1,i+1}&\cdots &a_{1n}\\a_{21}&a_{22}&\cdots &a_{2,i-1}&b_{2}&a_{2,i+1}&\cdots &a_{2n}\\\vdots &\vdots &\ddots &\cdots &\cdots &\cdots &\ddots &\vdots \\a_{n1}&a_{n2}&\cdots &a_{n,i-1}&b_{n}&a_{n,i+1}&\cdots &a_{nn}\end{pmatrix}}}
Pokud je determinant matice soustavy nenulový,
det
A
≠
0
{\displaystyle \det \mathbf {A} \neq 0}
A
{\displaystyle \mathbf {A} }
regulární , pak má soustava právě jedno řešení, pro které platí
x
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det
A
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det
A
{\displaystyle x_{i}={\frac {\det \mathbf {A} _{i}}{\det \mathbf {A} }}}
pro
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{\displaystyle i=1,2,...,n}
x
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{\displaystyle x_{1}}
x
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{\displaystyle x_{n}}
det
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det
A
{\displaystyle {\frac {\det \mathbf {A} _{i}}{\det \mathbf {A} }}={\frac {\begin{vmatrix}a_{1,1}&\cdots &a_{1,i-1}&b_{1}&a_{1,i+1}&\cdots &a_{1,n}\\\vdots &\ddots &\vdots &\vdots &\vdots &\ddots &\vdots \\a_{j-1,1}&\cdots &a_{j-1,i-1}&b_{j-1}&a_{j-1,i+1}&\cdots &a_{j-1,n}\\a_{j,1}&\cdots &a_{j,i-1}&b_{j}&a_{j,i+1}&\cdots &a_{j,n}\\a_{j+1,1}&\cdots &a_{j+1,i-1}&b_{j+1}&a_{j+1,i+1}&\cdots &a_{j+1,n}\\\vdots &\ddots &\vdots &\vdots &\vdots &\ddots &\vdots \\a_{n,1}&\cdots &a_{n,i-1}&b_{n}&a_{n,i+1}&\cdots &a_{n,n}\\\end{vmatrix}}{\det \mathbf {A} }}=\sum _{j=1}^{n}b_{j}{\frac {\begin{vmatrix}a_{1,1}&\cdots &a_{1,i-1}&0&a_{1,i+1}&\cdots &a_{1,n}\\\vdots &\ddots &\vdots &\vdots &\vdots &\ddots &\vdots \\a_{j-1,1}&\cdots &a_{j-1,i-1}&0&a_{j-1,i+1}&\cdots &a_{j-1,n}\\a_{j,1}&\cdots &a_{j,i-1}&1&a_{j,i+1}&\cdots &a_{j,n}\\a_{j+1,1}&\cdots &a_{j+1,i-1}&0&a_{j+1,i+1}&\cdots &a_{j+1,n}\\\vdots &\ddots &\vdots &\vdots &\vdots &\ddots &\vdots \\a_{n,1}&\cdots &a_{n,i-1}&0&a_{n,i+1}&\cdots &a_{n,n}\\\end{vmatrix}}{\det \mathbf {A} }}}
Jestliže matici získanou vynecháním j-tého řádku a i-tého sloupce matice
A
{\displaystyle \mathbf {A} }
A
j
i
{\displaystyle \mathbf {A} _{ji}}
det
A
i
det
A
=
∑
j
=
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n
b
j
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+
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det
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det
A
{\displaystyle {\frac {\det \mathbf {A} _{i}}{\det \mathbf {A} }}=\sum _{j=1}^{n}b_{j}{\frac {(-1)^{i+j}\det \mathbf {A} _{ji}}{\det \mathbf {A} }}}
Zlomek ve výrazu je prvkem
(
A
−
1
)
i
,
j
{\displaystyle (\mathbf {A} ^{-1})_{i,j}}
inverzní matice
A
−
1
{\displaystyle \mathbf {A} ^{-1}}
det
A
i
det
A
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∑
j
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n
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b
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i
{\displaystyle {\frac {\det \mathbf {A} _{i}}{\det \mathbf {A} }}=\sum _{j=1}^{n}(\mathbf {A} ^{-1})_{i,j}b_{j}=(\mathbf {A} ^{-1}\mathbf {b} )_{i}}
Protože
A
x
=
b
{\displaystyle \mathbf {Ax} =\mathbf {b} }
det
A
≠
0
{\displaystyle \det \mathbf {A} \neq 0}
x
=
A
−
1
b
{\displaystyle \mathbf {x} =\mathbf {A} ^{-1}\mathbf {b} }
x
i
=
det
A
i
det
A
{\displaystyle x_{i}={\frac {\det \mathbf {A} _{i}}{\det \mathbf {A} }}}
Obrázky, zvuky či videa k tématu Cramerovo pravidlo ve Wikimedia Commons 