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- Determinant 2x2
- Determinant 3x3
- Determinant 3x3 symbolic
- Determinant 4x4
- Determinant 4x4 symbolic
- Determinant 5x5
- Determinant NxN

### Determinant of a 3x3 matrix according to the Sarrus Rule

The determinant is calculated as follows by the Sarrus Rule. Schematically, the first two columns of the determinant are repeated so that the major and minor diagonals can be virtual connected by a linear line. Then one makes the products of the main diagonal elements and adds this products. With the secondary diagonals you shall do the same. The difference between the two gives the determinant of the matrix.

$\mathrm{det\; A}=\left|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right|\phantom{\rule{0.5em}{0ex}}\left.\begin{array}{cc}{a}_{11}& {a}_{12}\\ {a}_{21}& {a}_{22}\\ {a}_{31}& {a}_{32}\end{array}\right|$

$\begin{array}{c}\phantom{\rule{0.3em}{0ex}}\\ ={a}_{11}{a}_{22}{a}_{33}+{a}_{12}{a}_{23}{a}_{31}+{a}_{33}{a}_{21}{a}_{32}\\ \phantom{\rule{0.3em}{0ex}}\end{array}$

$\begin{array}{c}\phantom{\rule{0.3em}{0ex}}\\ -\left({a}_{31}{a}_{22}{a}_{13}+{a}_{32}{a}_{23}{a}_{11}+{a}_{33}{a}_{21}{a}_{12}\right)\\ \phantom{\rule{0.3em}{0ex}}\end{array}$

### Laplace Expansion Theorem

The Laplace expansion theorem provides a method for calculating the determinant, wherein the determinant is developed according to a row or column. The dimension is reduced and can be gradually reduced more and more up to the scalar.

### Example of the Laplace expansion according to the first row on a 3x3 Matrix.

$\mathrm{det\; A}=\left|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right|$

The first element is given by the factor a_{11} and the sub-determinant consisting of the elements with green background.

$\left|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right|=>{a}_{11}\left|\begin{array}{cc}{a}_{22}& {a}_{23}\\ {a}_{32}& {a}_{33}\end{array}\right|$

The second element is given by the factor a_{12} and the sub-determinant consisting of the elements with green background.

$\left|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right|=>{a}_{12}\left|\begin{array}{cc}{a}_{21}& {a}_{23}\\ {a}_{31}& {a}_{33}\end{array}\right|$

The third element is given by the factor a_{13} and the sub-determinant consisting of the elements with green background.

$\left|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right|=>{a}_{13}\left|\begin{array}{cc}{a}_{21}& {a}_{22}\\ {a}_{31}& {a}_{32}\end{array}\right|$

With the three elements the determinant can be written as a sum of 2x2 determinants.

$\mathrm{det\; A}=\left|\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\\ {a}_{21}& {a}_{22}& {a}_{23}\\ {a}_{31}& {a}_{32}& {a}_{33}\end{array}\right|$

$={a}_{11}\left|\begin{array}{cc}{a}_{22}& {a}_{23}\\ {a}_{32}& {a}_{33}\end{array}\right|$

$-{a}_{12}\left|\begin{array}{cc}{a}_{21}& {a}_{23}\\ {a}_{31}& {a}_{33}\end{array}\right|$

$+{a}_{13}\left|\begin{array}{cc}{a}_{21}& {a}_{22}\\ {a}_{31}& {a}_{32}\end{array}\right|$

It is important to consider that the sign of the elements alternate in the following manner.

$\left|\begin{array}{ccc}+& -& +\\ -& +& -\\ +& -& +\end{array}\right|$