Linear Algebra-Determinants and Eliminations

Its also possible to find Determinants through the use of Gauss-Jordan eliminations. Assuming the matrix is invertible and row reducible we can equate the row operations to the determinant with the following rules.
For each row division by a nonzero scalar k, we multiply those k's together.
For each row swap operation, we multiply by (-1)
No change due to row elimination.
Example:
|0 7 5 3 |
|1 1 2 1 |
|1 1 2 -1|
|1 1 1 2 |

|1 1 2 1 |
|0 7 5 3 | (-1)
|1 1 2 -1|
|1 1 1 2 |


|1 1 2 1 |
|0 7 5 3 |
|0 0 0 -2|
|1 1 1 2 |


|1 1 2 1 |
|0 7 5 3 |
|1 1 1 2 | (-1)
|0 0 0 -2|

|1 1 2 1 |
|0 7 5 3 |
|0 0 -1 1|
|0 0 0 -2|

Now we could complete the row reduction but we know from here that only two operations will occur. Dividing a row through by the first entry, and then starting from the last row elimination the above rows through multiplication and addition. The eliminations do not affect the result of the determinant, and all we have to do is multiply by k where k is the scalar value that we used to divide the row to get to a leading 1. Thus we can simply multiple the diagonal once the matrix has been reduced to an upper right triangular matrix.

det=(-1) (-1) (1 * 7 * -1 * -2)

det=14