|1 o |
| 0 2 |
= (2x1)-(0x0)=2
|1 -2|
|-1 3|
= (1x3)-(-2x-1)=1
| 2 -3 -2|
| -1 3 4 |
| -7 -2 8 |
2| 3 4 | X 3 | -1 4 | x -2 | -1 3|
|-2 8| | -7 8 | |-7 -2|
= 2(8-24)+3(8+28)-2(2+21)
= 94
did i do these right ?
|1 o |
| 0 2 |
= (2x1)-(0x0)=2
|1 -2|
|-1 3|
= (1x3)-(-2x-1)=1
| 2 -3 -2|
| -1 3 4 |
| -7 -2 8 |
2| 3 4 | X 3 | -1 4 | x -2 | -1 3|
|-2 8| | -7 8 | |-7 -2|
= 2(8-24)+3(8+28)-2(2+21)
= 94
did i do these right ?
there are 2 ways to evaluate a 3x3 determinant:
way #1) expansion by minors:
where is the 2x2 matrix obtained by removing row i and column j.
way #2) the rule of sarrus:
(often displayed as three diagonal products each way that "wrap around")
let us calculate:
using both methods.
the first way, we have a choice: expansion by a row, or by a column (we could use ANY row or column, but we'll use the first one to keep it simple). first we'll expand by the first row:
to do this we need to find the 2x2 determinants, first.
thus .
now we'll expand by the first column. we need to calculate 2 more 2x2 determinants (we'll re-use the one obtained by eliminating row 1 and column one):
thus .
finally, we'll use the rule of sarrus:
.
personally, i like the rule of sarrus, but expansion by minors generalizes better to larger matrices.