Let’s evaluate another determinant, now we want to use as many of our simplification steps as possible to make this easier. We have a pretty complicated looking determinant here. I think the first thing I’m going to do is try to get another zero in the top row, so that when I expand I’ve got a lot of zeros in that row.
What I’m going to do is, I’m going to multiply this first column by 4 and add the result to the third column. You can always take any column, multiply by a constant and add the result to another column. The same goes for rows. That step is not going to change the first two columns. I can just copy those down, 16 -9 30, 0 15 -20. But 4 times 16 is 64 plus -64 is 0. That’s my 0. 4 times -9 is -36 plus -12 is -48. 4 times 30 is 120 plus 10 is 130. I have my two zeros and expanding it across the top row is going to be a lot easier. But I still have a lot of common factors, in these rows that I can pull out.
Let me do that. I can actually pull a 16 out of the first row. That will leave 1, 0, 0. And this row, the second row looks like it’s got a factor of 3. So let me pull that out and that leaves -3 5 -16. Finally this bottom row has got a factor 10. So that’s going to be 3, -2 and 13. This is going to be much easier now.
What I have here on the left is just a bunch of constants. 48 times 10, 480, and then if I expand this determinant along the top row, I have 1 times this determinant. 1 times 5 -16, -2 13. And then I have minus zero times some other determinant, doesn’t even matter because it’s zero. Plus zero times another determinant, again doesn’t matter because you have a zero in front. So these are just going to disappear, these terms.
And then you’re left with 480 times, and this is going to be 65 minus 32. That’s going to be 33. 480 times 33, which ends up being 15840. And that’s our answer.
So you see when you’re dealing with a complicated looking determinant, you really should do as many simplification steps as you can before hand. It makes the process a lot easier.