Let A be the 3-by-3 matrix 1, 4, minus 4; minus 2, minus 2, minus 4, 3, minus 3, minus 3. Rather than applying our formula (sum over permutations), we're going to apply type I row operations to put this matrix into echelon form; this won't change the determinant, so we can take the determinant afterwards (at which point it will be the product of diagonal entries).
To clear the first column, we do R_2 maps to R_2 + 2 R_1 and R_3 maps to R_3 minus 3 R_1 which gives: 1, 4, minus 4; 0, 6, minus 12; 0, minus 15, 15. To clear the minus 15 below the diagonal, we do R_3 maps to R_3 + 5 R_2 over 2, giving: 1, 4, minus 4; 0, 6, minus 12; 0, 0, minus 15. This is in echelon form. Therefore det A equals 1 times 6 times minus 15, which equals minus 90.
Example:
Let B be the 4-by-4 matrix 2, minus 3, minus 4; 2, minus 3, 2, 4; 2, minus 1, minus 4, minus 3; 2, minus 3, 4, 2. Let's clear the first column with R_2 maps to R_2 minus R_1, R_3 maps to R_2 minus R_1, R_4 maps to R_4 minus R_1, yielding: 2, minus 3, minus 1, 4; 0, 0, 3, 0; 0, 2, minus 3, minus 7; 0, 0, 5, minus 2. Now let's swap R_2 with R_3: 2 , minus 3 , minus 1 , 4 ; 0 , 2 , minus 3 , minus 7; 0 , 0 , 3 , 0 ; 0 , 0 , 5 , minus 2. The determinant just changed sign! We'll need to remember to stick an extra minus sign in at the end. To clear the final 5 below the diagonal, we do R_4 maps to R_3 minus 5 over 3 R_3: 2 , minus 3 , minus 1 , 4 ; 0 , 2 , minus 3 , minus 7 ; 0 , 0 , 3 , 0 ; 0 , 0 , 0 , minus 2, which is now in echelon form.
The determinant of B is therefore minus (2 times 2 times 3 times minus 2), which equals 24.