Find determinants using Sarrus Rule ...
`[[3,2,4],[2,0,2],[4,2,3]]`

Solution:

`A=`	3   2   4   2   0   2   4   2   3

Write first 2 columns of matrix to right of 3rd column, so we have total 5 columns.

`A=`	3   2   4   3   2   2   0   2   2   0   4   2   3   4   2

`A=`	3     2     4     3     2     2     0     2     2     0     4     2     3     4     2

Now, add products of diagonals going from top to bottom (blue lines) and subtract products of diagonals going from bottom to top (red lines).

`={3*0*3+2*2*4+4*2*2}-{4*0*4+2*2*3+3*2*2}`

`=(0+16+16)-(0+12+12)`

`=32-24`

`=8`

Method-2: Determinant by expanding cofactors

`|A|`

=

`3`   `2`   `4`   `2`   `0`   `2`   `4`   `2`   `3`

=

`3`   ×    `0`   `2`   `2`   `3`

`-2`   ×    `2`   `2`   `4`   `3`

`+4`   ×    `2`   `0`   `4`   `2`

`=3 xx (0 × 3 - 2 × 2) -2 xx (2 × 3 - 2 × 4) +4 xx (2 × 2 - 0 × 4)`

`=3 xx (0 -4) -2 xx (6 -8) +4 xx (4 +0)`

`=3 xx (-4) -2 xx (-2) +4 xx (4)`

`= -12 +4 +16`

`=8`

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