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Mathematics-Online course: Linear Algebra - Matrices - Determinants

Expansion of Determinants

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The determinant of an $ n\times n$-matrix $ A$ can be expanded by any row or column:

\begin{displaymath} \begin{array}{rcll} \operatorname{det} A &=& \sum\limits... ...{i,l} & (\text{expansion\ by\ column}\ l)\, , \end{array} \end{displaymath}

where $ \tilde A_{i,j}$ denotes the matrix we obtain by deleting the $ i$-th row and the $ j$-th column of $ A$.
(Authors: Burkhardt/Höllig/Hörner)
Since $ \operatorname{det}A = \operatorname{det} A^{\operatorname t}$ it is sufficient to consider the expansion along a column.

We decompose the first column into

$\displaystyle a_l = \sum\limits_{i=1}^n a_{i,l}e_i\,, $

then it follows from the multilinearity of the determinant

$\displaystyle \operatorname{det}A = \sum\limits_{i=1}^n a_{i,l} \operatorname{d... ...erbrace{ \left(a_1,\ldots,a_{l-1},e_i,a_{l+1},\ldots,a_n\right) }_{=B_i}\,. $

Now, consider matrix $ B_{i}$. By $ n-1$ interchanges of columns the first column becomes the last one and the other columns remain in the same order as before. In the same manner the $ i$-th row can become the last row by $ n-i$ interchanges of rows:

$\displaystyle \operatorname{det}B_{i} = (-1)^{n-l}(-1)^{n-i}\operatorname{det... ...i,1}& \cdots& a_{i,l-1}& a_{i,l+1}&\cdots &a_{i,n} & 1\\ \end{array}\right) $

Expanding the determinant on the right hand side in a sum over permutations, we have to take into consideration only permutations $ \iota \in S_n$ with $ \iota_n=n$ (for other permutations the addends vanish). So we consider only the permutations of $ S_{n-1}$ and, thus,

$\displaystyle \operatorname{det} \left(\begin{array}{rrrrrrr} & & & & & & 0\\... ...dots &a_{i,n} & 1\\ \end{array}\right)=\operatorname{det}\tilde{A}_{i,l}\,. $

Myltiplying the addends by $ 1=(-1)^{2l+2i-2n}$ yields the given expansion formula.

(Authors: Burkhardt/Höllig/Hörner)
Expanding the determinant of matrix

$\displaystyle A=\left(\begin{array}{rrr} 2 & 1 & 1\\ 1 & 2 & 1\\ 1 & 1 & 2 \end{array} \right) $

along the first row we obtain
$\displaystyle \operatorname{det}A$ $\displaystyle =$ $\displaystyle (-1)^2\cdot 2 \operatorname{det}\left(\begin{array}{rr}2 & 1\\ 1 & 2\end{array}\right)$  
  $\displaystyle +$ $\displaystyle (-1)^3\cdot 1 \operatorname{det}\left(\begin{array}{rr}1 & 1\\ 1 & 2\end{array}\right)$  
  $\displaystyle +$ $\displaystyle (-1)^4\cdot 1 \operatorname{det}\left(\begin{array}{rr}1 & 2\\ 1 & 1\end{array}\right)$  
  $\displaystyle =$ $\displaystyle 2(4-1)-1(2-1)+1(1-2)=6-1-1=4\,.$  

(Authors: Burkhardt/Höllig/Hörner)
  automatically generated 4/21/2005