SOLUTION:

METHOD 1:

We change the matrix to an upper triangular one.



Switching two rows require that we multiply the matrix by –1. Now that it is ‘upper’ triangular… we know that the determinant of a triangular matrix is the product of the elements on the main diagonal.






METHOD 2:

We do no row operations and take it on faith that all lower or ‘backwards’ triangular matrices of dimension 4 will give the correct solution if we apply the diagonal rule.



We can verify this by using the cofactor expansion:
(This might, arguably be called method 3)

It should be noted that this does not hold for every ‘backwards diagonal’ matrix in different dimensions. This 4 X 4 matrix holds because we see that to make it into an upper triangular matrix we switch Row 1 and 4, then 2 and 3 to make the product to be 24*(-1)*(-1)=24.

If we were taking the dimension of 2 or 3, only two rows need to be switched to make it upper triangular and the determinant becomes negative product. For dimension 4 and 5, 2 row exchanges are needed, which led to the determinant being a positive product. 6 and 7 requires 3 exchanges with a negative product. We now see the pattern that emerges. When we completely reverse the order of the rows, the number of interchanges of adjacent rows required for this is . This number is even if n or n-1 is divisible by 4, and odd otherwise.

1. There is a zero row in this matrix. This means that even if you were able to perform row reduction on the matrix, you would end up with a zero on the diagonal. In the past we would say that the rows were linearly dependent and noninvertable.
   this causes the determinant to be 0.