# Simplifying Determinants - Problem 3

The whole point of row operations when you’re solving a determinant, is to make the determinant easier to evaluate. That’s really important the bigger the determinant gets. Here we have a 4 by 4, so doing doing row by column operations is going to be super important here.

Now looking at this determinant, my goal is to get some 0's on the top so it’s easy to expand across the top row. One way I can do that is I can double the second column and add it to the fourth column. That would give me a zero right there.

Doing so is not going to change the first three columns. I could just copy those over. I get 1 0 2 -2, -1 1 1 2, 0 -1 0 1. And I get 2 times -1 plus 2 is 0. 2 times 1 is 2 plus 0 is 2, 2 times 1 is 2, plus -1 is 1 and 2 times 2 is 4 plus 1, 5. So this column is twice this one. 2 times C2 plus C4.

Let’s do that again. I want to try to get rid of this guy. So I want to add the first column to the second column. That will give me another zero and I’ll have three zeros in the row, which will be really good. Doing so is not going to change the first, third or fourth column, so I can just copy those down. 1 0 2 -2, 0 -1 0 1, 0 2 1 5.

So just adding column 1 to column 2. I get 0, 0 plus 1, 1. 2 plus 1, 3, -2 plus 2, 0. Zeros are always nice wherever they show up. So this was C1 plus C2.

Now I’m going to expand this across the top row. Remember the rule, when you’re expanding, you’re going to need to keep in mind whether to use a plus or minus sign according to this chart. Now when you make a chart, no matter what size the determinant is, always make sure you start with the plus sign in the upper left and then alternate just like a checker board. Alternate across the rows and down the columns so it’s plus minus, plus minus here and so on.

So, going across the top row, I get plus 1 times 1 -1 2, its minor, 3 0 1 and 0 1 5. Of course the other minor is not going to matter because I have 0's here. So I’m going to have minus zero plus zero minus zero. Everything is going to end up zeroing out for these three terms. I’ll be left with the 3 by 3, 1 -1 2, 3 0 1, and 0 1 5. I can even make this determinant easier by doing a row operation.

Let me get a zero right here, by adding -3 times the top row to the second row. I multiply the top by -3. I get -3 plus 3 is 0. Multiply by -3 I get 3, plus 0 is 3. Multiply by -3 I get -6, plus 1 is -5. The others stay the same. And now I’m just going to expand across the first row. 1 times its minor, it's -3 -5, 1 5 and then of course I’ve got 0. So minus 0 for this one, plus 0 for this one. Done. It’s 1 times 15 minus -5. It’s 15 plus 5, 20. That’s my answer.

So don’t forget, use your row and column operations to make your determinant simpler. It’s really important to do this, when you’re staring off with a 3 by 3, or 4 by 4 or heaven forbid a 5 by 5 determinant.

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