Help me show that the determinant of this 3x3 matrix is zero without actually computing it!?

Question

One of the rules that won't change the value of a determinant of a matrix is: if we add to each element of one row, k times the corresponding element of another row, where k is ANY number (same statement applies to columns as well)(similar to row reduction, but applies to columns as well). I seem to be changng the value of the determinant by doing this... the first matrix i have(whose determinant is 0) is, 1st row: 0 -1 4, 2nd row: 1 0 -1, 3rd row: 4 1 0. I added -4 times the 2nd column to the first column(k= -4), and get, first row: 4 -1 4 , second row: 1 0 -1 , third row: 0 1 0. If u compute the determinant it is still zero! Now by using the above stated rule again, I add the 3rd column to the 1st column, (this theoretically should not change the value of the determinant, we take k=1), so the new matrix is, 1st row: 8 -1 4, 2nd row: 0 0 -1, 3rd row: 0 1 0. If we now solve for the determinant, we can see that it is 8. Why does this happen? It should still be zero!

Answer 1

Your initial matrix is:

[0, -1, 4]
[1 0, -1]
[4, 1, 0]

The determinant of this matrix is 8, not zero. Consider:

det = 0*0*0 + (-1)*(-1)*4 + 4*1*1 - 4*0*4 - 1*(-1)*0 - 0*1*(-1)
= 0 + 4 + 4 - 0 - 0 - 0
= 8

Now, if you add -4 times the second column to the first column, you should obtain:

[4, -1, 4]
[1 0, -1]
[0, 1, 0]

Whose determinant is still 8, since:

det = 4*0*0 + (-1)*(-1)*0 + 4*1*1 - 0*0*4 - 1*(-1)*4 - 0*1*(-1)
= 0 + 0 + 4 - 0 + 4 - 0
= 8

So given that the first two matrices had a determinant of 8, not zero, is it really so surprising that the third matrix also had a determinant of 8?