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<math>\det(A^\mathrm{T}) = \det(A) \,</math>
:Maatriksi transponeerimisel [[determinant]] ei muutu.
=== Tõestus ===
Kui <math>A=
\begin{bmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
a_{21} & a_{22} & \ldots & a_{2n} \\
\ldots & \ldots & \ldots & \ldots \\
a_{m1} & a_{m2} & \ldots & a_{mn} \\
\end{bmatrix}</math>
, siis<br />
<math>
\det(A^\mathrm{T})=
\det\left(
\begin{bmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
a_{21} & a_{22} & \ldots & a_{2n} \\
\ldots & \ldots & \ldots & \ldots \\
a_{m1} & a_{m2} & \ldots & a_{mn} \\
\end{bmatrix}
^\mathrm{T}\right)
=
\det\left(
\begin{bmatrix}
a_{11} & a_{21} & \ldots & a_{m1} \\
a_{12} & a_{22} & \ldots & a_{m2} \\
\ldots & \ldots & \ldots & \ldots \\
a_{1n} & a_{2n} & \ldots & a_{mn} \\
\end{bmatrix}
\right)</math>
<math>
=
\left\{
\begin{matrix}
m<>n=>0\\
m=n=1 => a_{1,1}\\
m=n=2 => a_{1,1}*a_{2,2}-a_{1,2}*a_{2,1}\\
m=n>2 =>
\sum_{j=0}^{n-1}
\left(
\prod_{i=1}^m a_{1+\left(\left(i+j-1\right)\pmod m\right)i}
\right)
-
\sum_{j=n-1}^{0}
\left(
\prod_{i=1}^m a_{1+\left(\left(i+j-1\right)\pmod m\right)i}
\right)
\end{matrix}
\right.
</math>
<math>
\det(A)=
\det\left(
\begin{bmatrix}
a_{11} & a_{12} & \ldots & a_{1n} \\
a_{21} & a_{22} & \ldots & a_{2n} \\
\ldots & \ldots & \ldots & \ldots \\
a_{m1} & a_{m2} & \ldots & a_{mn} \\
\end{bmatrix}
\right)
</math>
<math>
=
\left\{
\begin{matrix}
m<>n=>0\\
m=n=1 => a_{1,1}\\
m=n=2 => a_{1,1}*a_{2,2}-a_{1,2}*a_{2,1}\\
m=n>2 =>
\sum_{j=0}^{n-1}
\left(
\prod_{i=1}^m a_{i1+\left(\left(i+j-1\right)\pmod m\right)}
\right)
-
\sum_{j=n-1}^{0}
\left(
\prod_{i=1}^m a_{i1+\left(\left(i+j-1\right)\pmod m\right)}
\right)
\end{matrix}
\right.
</math>
[[Determinant|Determinandi]] ridade ja veergude vahetamine muudab vaid [[kommutatiivsus|kommutatiivsust]], kuna aga liitmine ja korrutamine on [[maatriks]]i puhul kommutatiivsed, siis <math>\det(A^\mathrm{T}) = \det(A) \,</math> on võrdsed.
=== Rakendus ===
* Leiab suvalise maatriksi deteminandi lahenduskäigu ja vastuse: http://smartlabs.pl/en/edu/wyznacznik
== 6 ==
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