Current time:0:00Total duration:10:25
0 energy points
Video transcript
I have a matrix A. It is an n by n matrix. And let me just write its rows like this. Let me just write it as r1. We could call them row vectors maybe. r2, I'm not doing it too formally. This is just to save on writing. And then it has an ith row, ri, and then you can keep going . That's an i right there. Then it has a jth row, rj, and you keep going and you get to the nth throw. It has n rows and n columns. So you get to rn just like that. That is my matrix. Just to make sure you get what I'm saying, so if I have a kth r-sub-k is equal to ak1. Maybe I'll write it as a vector. ak2 all the way to akn. So this is just your standard representation. I wrote it this way because we're just going to be dealing with rows in this video and it makes our notation a little bit easier. Let me focus on these two rows right here. And let me define another matrix B that is also an n by n matrix. And it's identical to matrix A except for one row. So it's identical to matrix A except for one row. You have r1 just like that; it's the same as that one there. r2, keep going, go down to our ri, even that one's identical. But rj I've now replaced. I'm replacing rj with rj minus a scalar multiple of ri. Minus c times ri. So minus a scalar multiple of that. I've replaced rj with that. So this is equivalent to the row operations we do we did our Gaussian Elimination, or when we put things in reduced row echelon form. And everything else in this matrix is the same as A. It's all the way down to rn. This is our matrix B. So let's think about what the determinant of B is going to be equal to. I'll do it in blue. Well, you could immediately say that B is equivalent to-- Well, you can imagine two vectors. You can imagine two matrices. One matrix that look like this. One matrix that look like r1, r2, all the way down ri, all the way down to rj. And then you keep going down to rn. That's one matrix, which you may have already noticed is identical to A. That's one matrix. Then you could have another matrix here that looks like this. It's identical everywhere. r1, r2, ri. Some dots there to show you I might have skipped some rows. Skip some more rows. And then you have c times times ri. c times ri. Let me do that in a different color. This is ri right here. And then you just keep going down to rn. Now, the determinant of B, you could view as the determinant of this guy. Let me write this here. The determinant of B is equal to the determinant of this guy plus the determinant of this guy. Hopefully, you remember a couple of videos ago, that if one matrix-- Let's have two matrices that are identical in every way except for one row. So these two matrices are completely identical except for what's going on on the jth row. Here you have a r-sub-j. Here you have a c times r-sub-i. So it's a scalar multiple of a row that you had up here, this guy. So this is ri, this is the ith row. Here you have an ri, here you have an ri. But here you have another version of r row, scalar multiple of ri, while here you have an rj. Now, if you have another matrix that is essentially identical to these two matrices, except for this one row. And in that one row, it looks like the addition of these two matrices-- and let me put a negative here. So if you kept this matrix completely identical, but if you were to replace it with the sum of these two rows. So rj minus c times ri, you'll get this matrix right here. You'll get matrix B. And we learned that the determinant of B is equal to the determinant of this guy and that guy. Remember, B is not the sum of these two matrices. B is identical to these two matrices, except for that one row where B's jth row is equivalent to the jth row of this guy, plus the jth row of that guy. And when I talk about adding rows, you're just adding their corresponding elements. So I could rewrite this so this row would look like-- The first term would be aj1 minus c times ai1. That would be the first term in that row. The second term of that row would be aj2 minus c times ai2. And it would go all the way to ajn minus ca-sub-in, the nth column. So that's all it means by that. So the determinant of B is equal to the determinant of this plus the determinant of this. The determinant of this, this thing right here is our matrix A. This is going to be the determinant of A. And what's the determinant of this? Well, let's break this down a little bit more. The determine of this is equal to what? This is completely equivalent to A, except one of its rows-- Sorry, this is completely equivalent to this matrix. Not equivalent to A. Be very careful. Don't listen to everything I say. It's not equivalent to A. The difference is that A has an rj here. This guy has a minus c times ri. So this is equivalent to this matrix. It's completely equivalent to this matrix right here. Let me do it like this. So you have an r1, r2, keep going, and you have an ri, then you have another ri. Let me clean this up a little bit. Let me clear this out just so I have some space to work with. You have an ri. You have that ri there. Then you have another ri. You have another ri right there. You have another ri. So the jth row has an ri there. Then you keep going and you have an r-sub-n. These two guys are completely equivalent except for this guy has a minus c times the jth row. Right? That's what this was, right here. This is the jth row. Everything we're doing is in the jth row. This has a minus c times the jth row. So the determinant of this guy right here-- Let me just be clear that I'm only taking the determinant of this guy right here. It's going to be equal to minus c times the determinant of-- let me write it this way --minus c times the determinant of r1, r2. You have your first ri. And then in the jth row you have another version of the ri. And then you go down to r-sub-n. So times that determinant. This is just the determinant of this. I've added brackets and straight lines. And we saw this a couple of videos ago. If you have a matrix, you just multiply one of its rows by a scalar, in this case minus c. It's equivalent to minus c. The determinant of the new matrix is equal to minus c times the determinant of your matrix. That's all I'm saying right here. But what is the determinant of this matrix? You might have already noticed that it has duplicate rows. It has an ri, and then in the ith row, then it has another ri in the jth row. Remember, we kind of decomposed this B matrix right here as the sum of-- Or its determinant can be described as the determinant of the sum of these two things. B isn't the sum of these two things. Every other element is identical to every other element in each of these guys. But this guy right here, he has duplicate ri's. And what do we know about the determinant of a matrix with duplicate entries? The determinant is zero. So this entry right here is zero. Minus c times 0, 0. So the determinant of this whole thing is 0. So the big take-away right here is that the determinant of B is equal to just the determinant of this thing, which was the determinant of A. This is a very big take-away. It's going to make our life very easy. The determinant of B is equal to the determinant of A. So if you start with some matrix, and you replace the jth row in this example, but any row. If you replace any row with that row minus some scalar multiple of another row-- we picked ri in this case, that would be ri --the determinant will not be changed. You have to be very particular about how you say it because, obviously, if you just multiplied something by a scalar-- if you were to change its determinant, or if you do other things. If you just take a row, if you take the jth row, and you replace it with the jth row minus c times the ith row times some other row, which is equivalent to just a row operation that we have been doing, then it will not change your determinant. Which is a very big take-away because now we can carefully do some row operations and know that the determinant will not change.