Determinants 2

Sarrus' Rule is a formula used to find the determinant of a 3 by 3 matrix. Here is how it works, given a matrix

$\left( \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right)$

The numbers represent the row and column of the entry in the matrix

Copy the first two columns to the end of the matrix, so you will get

$\left( \begin{array}{ccc} a_{11} & a_{12} & a_{13} & a_{11} & a_{12}\\ a_{21} & a_{22} & a_{23} & a_{21} & a_{22}\\ a_{31} & a_{32} & a_{33} & a_{31} & a_{32}\end{array} \right)$

Draw diagonals of 3 and multiply the entries together.

We get $a_{11}a_{22}a_{33}$ , $a_{12}a_{23}a_{31}$ , $a_{13}a_{21}a_{32}$ with diagonals sloping down and

$a_{12}a_{21}a_{33}$ , $a_{11}a_{23}a_{32}$ and $a_{13}a_{22}a_{31}$ sloping up.

Lastly, we add the diagonals sloping down and minus the diagonals sloping up to get the formula

$a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{12}a_{21}a_{33}-a_{11}a_{23}a_{32}-a_{13}a_{22}a_{32}$

We apply this into our number-padded matrix and we get $1 \times 5 \times 9 + 2 \times 6 \times 8 + 3 \times 4 \times 8 - 2 \times 4 \times 9 - 1 \times 6 \times 8 - 3 \times 5 \times 7 = 0$