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Mathematics-Online course: Linear Algebra - Matrices - Determinants

Transformation of Determinants

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The determinant of a matrix having two identical rows or columns equals zero. From this it follows that the determinant does not change if we add a multiple of a column (row) to another column (row).
(Authors: Burkhardt/Höllig/Hörner)
It is sufficient to prove the statement for columns, since by transposing a matrix rows become columns and vice versa, and since $ \operatorname{det}A=\operatorname{det}A^{\operatorname t}$.

Interchanging two columns the determinant changes the sign by definition. Hence, interchanging the two identical columns the matrix does not change and, thus, the determinant equals zero.

If we add the $ \lambda$-fold of column $ a_j$ to column $ a_i$, $ i\neq j$, then we obtain the matrix

$\displaystyle A'=\left(a_1,\ldots,a_i+\lambda a_j,\ldots,a_j,\ldots,a_n\right)\,. $

By linearity we can decompose the determinant into

$\displaystyle \operatorname{det}A' = \operatorname{det}\left(a_1,\ldots,a_i,\... ...\lambda\operatorname{det}\left(a_1,\ldots,a_j,\ldots,a_j,\ldots,a_n\right)\,. $

The second matrix contains two identical columns and, hence, its determinant equals zero.

(Authors: Burkhardt/Höllig/Hörner)
By adding multiples of rows of a matrix to other rows and interchanging rows or columns the matrix can be transformed into an upper triangle matrix the determinant of which can be obtained as product of all diagonal entries. Each interchange of two rows or two columns leads to a change of sign.

For example, the determinant of

$\displaystyle A=\left(\begin{array}{rrrr} 1 & 4 & 0 & 2\\ 0 & 8 & 0 & 4\\ 3 & 3 & 3 & 2\\ 0 & 6 & 0 & 4 \end{array} \right) $

can be calculated as follows.

row 3 - $ 3\cdot$ row 1:

$\displaystyle \operatorname{det}A= \operatorname{det} \left(\begin{array}{rrr... ...\ 0 & 8 & 0 & 4\\ 0 & -9 & 3 & -4\\ 0 & 6 & 0 & 4 \end{array} \right) $

Interchange of column 2 and column 4:

$\displaystyle \operatorname{det}A= -\operatorname{det} \left(\begin{array}{rr... ...\ 0 & 4 & 0 & 8\\ 0 & -4 & 3 & -9\\ 0 & 4 & 0 & 6 \end{array} \right) $

row 3 + row 2, row 4 - row 2:

$\displaystyle \operatorname{det}A= -\operatorname{det} \left(\begin{array}{rr... ...& -1\\ 0 & 0 & 0 & -2 \end{array} \right)=-(1\cdot4\cdot3\cdot(-2))=24\,. $

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