The online calculator calculates the value of the determinant of a 3x3 matrix after Sarrus rule and with the Laplace expansion in a row or column.

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

For a 3x3 matrix the determinat can be calculated with the Sarrus rule. The Sarrus rule uses the diagonals for the calculation. The calculator shows the calculation steps. For illustration the elements of the main diagonals are colored in green and the elements of the secondary diagonals are colored in blue. In gray the first two columns are repeated for easier reading of the diagonals.

A general method to calculate the determinat is given by the Laplace expansion theorem. The theorem can be used from any row or column. The calculator shows the expansion for a selected row or column. You can select the row or column to be used for expansion.

Note:

If leading coefficients zero then should be columns or rows are swapped accordingly so that a divison by the leading coefficient is possible. The value of the determinant is correct if, after the transformations the lower triangular matrix is zero, and the elements of the main diagonal are all equal to 1.

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.

The Laplacian development theorem provides a method for calculating the determinant, in which the determinant is developed after a row or column. The dimension is reduced and can be reduced further step by step up to a scalar.

$\mathrm{det\; A}=\sum _{i=1}^{n}{-1}^{i+j}\cdot {a}_{ij}\mathrm{det}{A}_{ij}\text{( Expansion on the j-th column )}$

$\mathrm{det\; A}=\sum _{j=1}^{n}{-1}^{i+j}\cdot {a}_{ij}\mathrm{det}{A}_{ij}\text{( Expansion on the i-th row )}$

where A_{ij}, the sub-matrix of A, which arises when the i-th row and the j-th column are removed.

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

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.

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

With the Gauss method, the determinant is so transformed that the elements of the lower triangle matrix become zero. To do this, you use the row-factor rules and the addition of rows. The addition of rows does not change the value of the determinate. Factors of a row must be considered as multipliers before the determinat. If the determinat is triangular and the main diagonal elements are equal to one, the factor before the determinant corresponds to the value of the determinant itself.

$\mathrm{det\; A}=\left|\begin{array}{cccc}{a}_{11}& {a}_{12}& \dots & {a}_{1n}\\ {a}_{j1}& \phantom{\rule{0.3em}{0ex}}{a}_{j2}& \dots & \phantom{\rule{0.3em}{0ex}}{a}_{jn}\\ & \vdots \\ {a}_{n1}& {a}_{n2}& \dots & {a}_{nn}\end{array}\right|={\lambda}\phantom{\rule{0.3em}{0ex}}\left|\begin{array}{cccc}1& {a}_{12}& \dots & {a}_{1n}\\ 0& 1& \dots & {a}_{jn}\\ & \vdots \\ 0& 0& \dots & 1\end{array}\right|={\lambda}\phantom{\rule{0.3em}{0ex}}\mathrm{det\; A\text{'}}={\lambda}$

Here is a list of of further useful calculators:

Index Matrix Determinant Determinant 2x2 Determinant 3x3 Determinant 3x3 symbolic Determinant 4x4 Determinant 4x4 symbolic Determinant 5x5 Determinant 5x5 symbolic Determinant NxN