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n x n determinant

Defining the determinant for nxn matrices. An example of a 4x4 determinant. Created by Sal Khan.

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  • mr pants teal style avatar for user aclarke.clarke758
    Is it possible to find the determinant of an mxn matrix? Or is the determinant only defined for an nxn matrix?
    (11 votes)
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  • blobby green style avatar for user prionresearcher
    Why did u make the ex. matrix have at have straight lines like | | rather than the matrix border like [ ]?
    (0 votes)
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    • blobby green style avatar for user briana.harder
      He's trying to find the determinant of the matrix, which is written with the || lines instead of the [] brackets for the matrix. If you go back to the video on the determinants of 2x2 matrices, Sal talks about the various notations used for the determinant.
      (41 votes)
  • leafers ultimate style avatar for user Navid
    The computations done here is like taking a cross product. Does this connection have any significance?
    (8 votes)
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    • leaf green style avatar for user SteveSargentJr
      Yes, it sure does! In fact, for people that are familiar with determinants, there is an easy way to remember how to take the cross product:

      First, given two vectors a & b in R^3.
      Next, the Cross product
      a x b
      is obtained by taking the determinant of the matrix C, where:

      |C| =
      | i j k |
      |a1 a2 a3|
      |b1 b2 b3|

      [Note: I'm having some trouble typing out this matrix--it's supposed to be a 3x3 square matrix]

      i.e. the elements in the first row are the unit vectors in the x, y, & z directions, respectively, and the second row is the vector a and the third row is the vector b.

      The main difference is that instead of ending up with a single number (as you normally do when calculating a determinant), you end up with a vector (because of the unit vectors in the top row).
      I personally have always had trouble remembering the formula for the Cross Product of two vectors and so this different approach has been really helpful.

      Hope this helps!
      (18 votes)
  • winston default style avatar for user thomasinthailand
    I think that the 2 by 2 matrix determinant can be defined by the recursion as well, if we make the determinant of a 1x1 matrix just the value of the entry in the matrix. Is that right?
    (12 votes)
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    • male robot hal style avatar for user Cameron
      yes, a determinant for a 1x1 matrix is itself i.e. det([x])=x
      so for a 2x2 matrix
      det( [[a b] , [c d]] ) = a*det([d]) - b*(det([c]) =ad-bc
      it makes sense that a 1x1 matrix has a determinant equal to itself,
      because[a] [x] = [y] , or
      ax=y
      this is easily solvable as x=y/a, but the solution for x is undefined when
      a=0=det([a])
      (7 votes)
  • blobby green style avatar for user Jordan Kelley
    Is it correct to say the number of Non-invertible transformations are less 'dense'(in a loose and not rigorous sense) than invertible transformations? Considering the number of solutions where the determinant is equal to zero is less the the number of solutions where the determinants are equal to anything else, in fact the ratio of non-invertible transformations to invertible transformations is infinitesimal, correct?
    (5 votes)
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  • aqualine seedling style avatar for user David Kyle
    it seems to me that calculating the determinant quite often would take more time computationally than just doing the operations to get the rref of the matrix in question if you were just trying to find out if a matrix is invertible since at the very first non-pivot column or zeroed row you could stop operations.

    Furthermore if you take the effort to get the rref of a matrix it only doubles the number of operations to actually get the inverted matrix at the end which seems to be valuable information if you are interested in if the matrix was invertible in the first place.

    it just seems to me that the determinant might not be very valuable information in application. Is there some part of this that I'm missing? Is it actually a significant reduction in computational time to find inverses with determinants? I could see it possibly being computationally less time if we know ahead of time that the matrix is invertible to begin with. But how much so? I've been told that finding the rref is approximately O(1/3n^3). What is the determinate?
    (2 votes)
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    • blobby green style avatar for user Kjeld Schmidt
      The algortihm shown here, Laplace's Algortihm for finding determinants, has a horrendous O(n!). Another propular algorithm, LU-Decomposition is a mere O(n^3), so it's much, much better.

      Additionally the determinant does many things beside telling you whether or not theres an inverse.
      (3 votes)
  • mr pants teal style avatar for user David.V.Mireles
    How do I know that if the determinant of this general matrix is 0, then the matrix is not invertible?
    (2 votes)
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    • male robot hal style avatar for user Yamanqui García Rosales
      Remember that for a matrix to be invertible it's reduced echelon form must be that of the identity matrix.

      When we put this matrix in reduced echelon form, we found that one of the steps was to divide each member of the matrix by the determinant, so if the determinant is 0, we cannot do that division, and therefore we cannot put the matrix in the form of the identity matrix, and therefore the matrix is not invertible.
      (2 votes)
  • blobby green style avatar for user http://facebookid.khanacademy.org/100000686238310
    why don't you show laplace method?
    (2 votes)
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  • male robot hal style avatar for user Wudaifu
    At , should Sal have said & written +/- a sub1n times the DETERMINANT of A sub 1n instead of just a sub1n times A sub1n ?
    (2 votes)
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  • aqualine ultimate style avatar for user Adit Mistry
    This question isn't exactly related to this video, but I'd be grateful if you could help. I saw this one theorem where It said:

    If A is an n x n matrix, then the following are equvalent:
    ( 1 ) det A cannot be equal to 0
    ( 2 ) rank (A) = n
    ( 3 ) A is invertible.

    I don't understand the first condition. Isn't it true that is 2 rows are the same in a matrix, then the determinant is 0?

    e.x.:
    | 1 2 3 |
    | 0 1 4 |
    | 0 1 4 |

    This is an n x n matrix right? and the det is 0 right?
    (1 vote)
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    • primosaur ultimate style avatar for user Derek M.
      Yes, that is an nxn matrix. The theorem is not saying that every nxn matrix has non zero determinant, it's saying that an nxn matrix is invertible if and only if the determinant is not 0.

      You found an nxn matrix with determinant 0, and so the theorem guarantees that this matrix is not invertible.

      What "the following are equivalent" means, is that each condition (1), (2), and (3) mathematically mean the same thing. It is not saying that every nxn matrix has a nonzero determinant. The theorem says if a matrix is nxn, then conditions (1), (2), and (3) mathematically mean the same thing.
      (3 votes)

Video transcript

So far we've been able to define the determinant for a 2-by-2 matrix. This was our definition right here: ad minus bc. And then we were able to broaden that a bit by creating a definition for the determinant of a 3-by-3 matrix, and we did that right here, where we essentially said the determinant is equal to each of these terms-- you could call these maybe the coefficient terms-- times the determinant of the matrix-- you can kind of view it as the submatrix produced-- when you get rid of each of these guys' column and row. So when you got rid of this guy's column and row, you're left with this matrix. So we said this guy times the determinant of this. And we kept switching signs, minus this guy times the determinant, if you move his column and his row. So it was left with these terms right there to get that determinant. Then finally, you switched signs again. So plus this guy times the determinant of the 2-by-2 matrix if you get rid of this row and this column. So this thing right here, which was this matrix. Now let's see if we can extend this to a general n-by-n matrix. So let's write out our n-by-n matrix right over here. I'll do it in blue. So let's say I have some matrix A that is an n-by-n matrix, so it's going to look like this. This would be a11, that would be a12, and we would go all the way to-- you're going to have n columns, a1n. And when you go down, this is going to be your second row: a21, and it's going to go all the way down to an1, because you have n rows as well. And then if you go down the diagonal all the way, this right here would be ann. So there is my n-by-n matrix. Now, before I define how to find the determinant of this, let me make another definition. Let me define-- so this is my matrix A. Let me define a submatrix Aij to be equal to-- see this is n by n, right? So this is going to be an n minus 1 by n minus 1 matrix. So if this is 7 by 7, the submatrix is going to be 6 by 6, one less in each direction. So this is going to be the n minus 1 by n minus 1 matrix you get if you essentially ignore or if you take away-- maybe I should say take away. Let's say ignore, like the word ignore. If you ignore the i-th row, this right here is the row, the i-th row and the j-th column of a. So, for example, let's go back to our 3 by 3 right here. This thing could be denoted based on that definition I-- we could have called this, this was a11, this term right here. We could denote the matrix when you get rid of the first column and the first row or the first row and the first column, we could call this thing right here, we could call that big matrix A11. So this was big matrix A11. This is big matrix A21, or actually, this matrix was called C, so this would be C11 right there. We could call this one, this would be matrix C12. Why is that? Because if you get rid of the first row, let me get rid of the first row, right? The first term is your row. If you get rid of the first row and the second column, this is the matrix that's left over: 2, 3, 4, 1. So this is this guy and this guy. 2, 3, 4, 1. So this is the submatrix c because this is the big matrix C, But this one is C12. I know it's a little bit messy there. So that's all we mean by the submatrix. Very similar to what we did in the 3 by 3 case. You essentially get rid of-- so if you want to find out this guy's submatrix, you would call that a11, and you would literally cross out the first row and the first column, and everything left over here would be that submatrix. Now, with that out of the way, we can create a definition, and it might seem a little circular to you at first, and on some level it is. We're going to define the determinant of A to be equal to-- this is interesting. It's actually a recursive definition. I'll talk about that in a second. It's equal to-- we start with a plus. It's equal to a11 times the submatrix if you remove this guy's row and column. So that, by definition, is just A, big capital A11's determinant. So that's exactly what we did. Let me write that a little bit neater. So times the determinant of its submatrix, so the determinant of A11. So you take A11, you get rid of its column and its row or its row and its column, and everything else, you find the determinant of that. Actually, let me write it in terms of-- let me write it this way. a11 times the determinant of the submatrix A11. And then we switch sides. We're just going to go along this row, and then you do minus a12 times the determinant of its submatrix, which we'll just call A12. We would get rid of this row and this column, and everything left would be this matrix A12. We want to find its determinant. And then we'll take the next guy over here would be a13. So we switch signs with minus. Now, you go plus, so a13 times the determinant of its submatrix. So if this is n by n, these each are going to be n minus 1 by n minus 1. So the determinant of A13. And you're just going to keep doing that, keep switching signs, so it's going to be a minus and then a plus and you keep going all the way-- and then I don't know. It depends on whether an, whether we're dealing with an odd number or an even number. If we're dealing with an even number, this is going to be a minus sign. If it's an odd number, it's going to be a plus sign, but you get the idea. It's either going to be a plus or a minus, not just-- if it's odd, this is going to be a plus. If it's an even n, it's going to be a minus, All the way to a1n, the n-th column times its submatrix, A1n. With that submatrix, you get rid of the first row and the n-th column, and it's going to be everything that's left in between. And you immediately might say, Sal, what kind of a definition is this? You defined a determinant for an arbitrary n-by-n matrix in terms of another definition of a determinant. How does this work? And the reason why this works is because the determinant that you use in the definition are determinants of a smaller matrix. So this is a determinant of an n minus 1 by n minus 1 matrix. And you're saying hey, Sal, that still doesn't make any sense because we don't know how to find the determinant of an n minus 1 by n minus 1 matrix. Well, you apply this definition again, and then it's going to be in terms of n minus 2 times n-- or n minus 2 by n minus 2 matrices. And you're like how do you do that? Well, you keep doing it, and you're going to get all the way down to a 2-by-2 matrix. And that one we defined well. We defined the determinant of a 2-by-2 matrix not in terms of a determinant. We just defined it in terms of a times-- we defined it as-- let me write it up here. It was a times d minus b times c. And you can see. I mean, we could just go down to the 3 by 3, but the 2 by 2 is really the most fundamental definition. And you could see that the definition of a 3-by-3 determinant is a special case of the general case for an n by n. We take this guy and we multiply him times the determinant of his submatrix right there. Then we take this guy where we switch signs. We have a minus. And we multiply him times the determinant of his submatrix, which is that right there. Then you do a plus. You switch signs and then you multiply this guy times the determinant of his submatrix, which is that right there. So this is a general case of what I just defined. But we know it's never that satisfying to deal in the abstract or the generalities. We want to do a specific case. And actually, before I do that, let me just introduce a term to you. This is called a recursive formula. And if you become a computer science major, you'll see this a lot. But a recursive function or a recursive formula is defined in terms of itself. But the things that you use in the definition use a slightly simpler version of it, and as you keep going through, or you keep recursing through it, you get simpler and simpler versions of it until you get to some type of base case. In this case, our base case is the case of a 2-by-2 matrix. You keep doing this, and eventually you'll get to a determinant of a 2-by-2 matrix, and we know how to find those. So this is a recursive definition. But let's actually apply it because I think that's what actually makes things concrete. So let's take-- this is going to be computationally intensive, but I think if we focus, we can get there. So I'm going to have a 4-by-4 matrix: 1, 2, 3, 4. 1-- throw some zeroes in there to make the computation a little bit simpler, 0, 1, 2, 3, and then 2, 3, 0, 0. So let's figure out this determinant right there. This is the determinant of the matrix. If I put some brackets there that would have been the matrix. But let's find the determinant of this matrix. So this is going to be equal to-- by our definition, it's going to be equal to 1 times the determinant of this matrix right here if you get rid of this row and this column. So it's going to be 1 times the determinant of 0, 2, 0; 1, 2, 3; 3, 0, 0. That's just this guy right here, this matrix right there. Then I'm going to have a 2, but I'm going to switch signs. So it's minus 2 times the determinant if I get rid of that row and this column, so it's 1, 2, 0. I'm ignoring the zero because it's in the same column as the 2: 1, 2, 0; 0, 2, 3, and then 2, 0, 0. And then I switch signs again. It was a minus, so now I go back to plus. So I do that guy, so plus 3 times the determinant of his submatrix. Get rid of that row and get rid of that columm, I get a 1, 0, 0. I get a 0, 1, 3. I skip this column every time. Then I get a 2, 3, 0, just like that. We're almost done. One more in this column. Let me switch to another color. I haven't used the blue in this yet. So then I'm going to do a minus 4. Remember, it's plus, minus, plus, minus 4 times the determinant of its submatrix. That's going to be that right there. So it's 1, 0, 2; 0, 1, 2; 2, 3, 0, just like that. And now we're down to the 3-by-3 case. We could use the definition of the 3 by 3, but we could just keep applying this recursive definition. So this is going to be equal to-- let me write it here. It's 1 times-- what's this determinant? This determinant's going to be 0 times the determinant of that submatrix, 2, 3, 0, 0. That was this one right here. And then we have minus 2, minus this 2-- remember, we switched signs-- plus, minus, plus, so minus 2 times its submatrix, so it's 1, 3, 3, 0. And then finally plus 0 times its submatrix, which is this thing right here: 1, 2, 3, 0, just like that. And then we have this next guy right here. As you can see, this can get a little bit tedious, but we'll keep our spirits up. So minus 2 times 1 times its submatrix, so that's this guy right here-- times the determinant of its submatrix 2, 3, 0, 0. Then minus 2 times-- get rid of that row, that column-- 0, 3, 2, 0. And then plus 0 times 0, 2, 2, 0. That's this one right there. Halfway there, at least for now. And then we get this next one, so we have a plus 3. Bring out our parentheses. And then we're going to have 1 times its sub-- I guess call it sub-determinant. So 1 times the determinant 1, 3, 3, 0, right? You get rid of this guy's row and column, you get this guy right there. And then minus 0-- get rid of this row and column-- times 0, 3, 2, 0. Then you have plus 0 times its sub-determinant 0, 1, 2, 3. Three-fourths of the way there. One last term. Let's hope we haven't made any careless mistakes. Minus 4 times 1 times 1, 2, 3, 0 right there. Minus 0 times-- get rid of those two guys-- 0, 2, 2, 0. And then plus 2 times 0, 1, 2, 3, right? Plus 2-- get rid of these guys-- 0, 1, 2, 3. Now, we've defined or we've calculated or we've defined our determinant of this matrix in terms of just a bunch of 2-by-2 matrices. So hopefully, you saw in this example that the recursion worked out. So let's actually find what this number is equal to. A determinant is always just going to be a number. So let me get a nice vibrant color. This is 0 times-- I don't care. 0 times anything's going to be 0. 0 times anything is going to be 0. 0 times anything's going to be 0. 0 times anything's going to be 0. Just simplifying it. These guys are 0 because it's 0 times that. 0 times this is going to be equal to 0. So what are we left with? This is going to be equal to 1 times-- this is all we have left here is a minus 2 times-- and what is this determinant? It's 1 times 0, which is 0. It's 0-- let me write this. This is going to be 1 times 0 is 0, minus 3 times 3 is 0 minus 9, so minus 9. This right here is just minus 9. So minus 2 times minus 9. That's our first thing, I'll simplify it in a second. Now let's do this term right here. So it's minus 2 times-- now what's this determinant? 2 times 0 minus 0 times 3. That's 0 minus 0. So this is 0. That guy became 0, so we can ignore that term. This term right here is 0 times 0, which is 0, minus 2 times 3. So it's minus 6. So it's minus 2 times-- so this is a minus 6 right here. You have a minus 2 times a minus 6, so that's a plus 12. So I'll just write a plus 12 here. This minus 2 is that minus 2 right there. And then we have a plus 3. And then this first term is 1 times 0, which is 0, minus-- let me make the parentheses here-- 1 times 0, which is 0, minus 3 times 3, which is minus 9 times 1. So it's minus 9. Everything else was a 0. We're in the home stretch. We have a minus 4. Let's see, this is 1 times 0, which is 0, minus 3 times 2, so minus 6. So this is minus 6 right here. Minus 6, this is 0, and then we have this guy right here. So we have 0 times 3, which is 0, minus 2 times 1. So that's minus 2, and then you have a minus 2 times a plus 2 is minus 4. So now we just have to make sure we do our arithmetic properly. This is 1 times plus 18, so this is 18, right? Minus 2 times minus 9. This right here is minus 24. This right here is minus 27. And then this right here, let's see, this is minus 10 right here. That is minus 10. Minus 4 times minus 10 is plus 40. Let's see if we can simplify this a little bit. If we simplify this a little bit-- I don't want to make a careless mistake right at the end. So 18 minus 24, 24 minus 18 is 6, so this is going to be equal to minus 6, right? 18 minus 24 is minus 6. And then-- let me do it in green-- now what's the difference? If we have minus 27 plus 40, that's 13, right? It's positive 13. So minus 6 plus positive 13 is equal to 7. And so we are done! After all of that computation, hopefully we haven't made a careless mistake. The determinant of this character right here is equal to 7. The determinant is equal to 7. And so the one useful takeaway, we know that this is invertible because it has a non-zero determinant. Hopefully, you found that useful.