Probably should mention that the Rule of Sarus only applies to 3x3. You cannot use it for 4x4 or higher.

Actually, there seems to be a method to extend this to 4x4 and even higher matrices. I found this one article, where it is described: http://amca01.wordpress.com/2012/08/16/sarrus-rule-and-extensions-to-higher-orders/ But I wouldn't use it, the recursive method is less error prone and easier to remember.

Since we solve for the determinate of a 4x4 matrix by essentially breaking it down into smaller 3x3 matrices, can we apply the Rule of Sarrus after manipulating a 4x4 into a sequence of 3x3 matrices?

Yes, that would be a valid approach. Note that if you had to find the determinant of a 4x4 or bigger matrix, the methods shown here do not scale well. The number of computations required grows a lot. A really nice thing to do is to row reduce the matrix to what is called an upper triangular (means all the entries below the main diagonal are zero). Then the determinant is simply the product of all the entries on the main diagonal. If you accessed this video from the Linear Algebra playlist, this method will be presented soon.

Would this work for a four by four or two by two? thanks

You don't really need any special shortcuts for a 2x2, but you may notice that finding the determinant for a 2x2 matrix is already kind of similar to the Rule of Sarrus.

I am just learning the Rule of Sarrus in my algebra 2 class and was asked by my teacher to find out why this works. Is there a reason or is it just something that works and we all do it? Thanks!

Well, many of us don't know why using "minors and cofactors" works. Does your teacher want to find out the answer to his question, or was it meant just as an exercise for you? If you accept that minors and cofactors is accurate, then Sal has proved Sarrus' rule in the video based on that (I think Sarrus' rule is only for 3x3 determinants). If you want to prove Sarrus' rule from definitions, then here's the definition of a determinant: "def.: determinant: 2 (mathematics): A square array of numbers bordered on the left and right by a vertical line and having a value equal to the algebraic sum of all possible products where the number of factors in each product is the same as the number of rows or columns, each factor in a given product is taken from a different row and column, and the sign of a product is positive or negative depending upon whether the number of permutations necessary to place the indices representing each factor's position in its row or column in the order of the natural numbers is odd or even." For the first product, aei = a11a22a33, there are no permutations, so this product is positive. For bfg = a12a23a31, the second subscripts are in the order (2, 3, 1) (the first subscripts are always in numerical order (no permutations) to prevent confusion). To get from (1, 2, 3) to (2, 3, 1) first switch 2 and 1 to get (2, 1, 3) then switch 1 and 3 to get (2, 3, 1). that's an even number of permutations (two), so bfg = a12a23a31 is positive. Similarly cdh is positive. For ceg = a13a22a31, to get from (1, 2, 3) to (3, 2, 1), switch first 1 and 2, then 1 and 3, to get (2, 3, 1), then switch 2 and 3 to get (3, 2, 1); which is 3 permutations (an odd number), so it's "-ceg". Similarly you get "-afh" and "-bdi". So now we have proven that all the terms are correct. Now you can sketch the pattern (A augmented by repeating its first two columns, with ovals around all the resulting diagonals) of the elements as shown in the video, which helps you remember the terms and their signs. \\

The outcome of this Det (A) = 45, while in the past video with the 3 x 3 matrix the outcome of the Det (A) was 35. How?

a(3,1) in the past video is 1, but here it's -1.

To find determinants, wouldn't it be easier just to use expansion by minors?

Yes, it sure can be. If you have a lot of zeros in the matrix it is a great way to do it by hand. In general I believe that there are faster ways to compute the determinant by factoring the matrix into the product of a permutation matrix, a diagonal matrix, and an upper triangular matrix.

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Course: Linear algebra > Unit 2

Lesson 5: Finding inverses and determinants

Rule of Sarrus of determinants

Name: Rule of Sarrus of determinants
Uploaded: 2011-02-20T16:46:46Z
Description: A alternative "short cut" for calculating 3x3 determinants (Rule of Sarrus)

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A alternative "short cut" for calculating 3x3 determinants (Rule of Sarrus). Created by Sal Khan.

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Maureen Hamilton
Posted 12 years ago. Direct link to Maureen Hamilton's post “Probably should mention t...”
Probably should mention that the Rule of Sarus only applies to 3x3. You cannot use it for 4x4 or higher.
Button navigates to signup pageComment on Maureen Hamilton's post “Probably should mention t...”
(55 votes)
Answer
- Norberto Galdini
  Posted 11 years ago. Direct link to Norberto Galdini's post “Actually, there seems to ...”
  Actually, there seems to be a method to extend this to 4x4 and even higher matrices. I found this one article, where it is described:
  http://amca01.wordpress.com/2012/08/16/sarrus-rule-and-extensions-to-higher-orders/
  
  But I wouldn't use it, the recursive method is less error prone and easier to remember.
  Button navigates to signup page
  (12 votes)
Jon Trossbach
Posted 12 years ago. Direct link to Jon Trossbach's post “Since we solve for the de...”
Since we solve for the determinate of a 4x4 matrix by essentially breaking it down into smaller 3x3 matrices, can we apply the Rule of Sarrus after manipulating a 4x4 into a sequence of 3x3 matrices?
Button navigates to signup pageButton navigates to signup page
(6 votes)
Answer
- newbarker
  Posted 12 years ago. Direct link to newbarker's post “Yes, that would be a vali...”
  Yes, that would be a valid approach.
  
  Note that if you had to find the determinant of a 4x4 or bigger matrix, the methods shown here do not scale well. The number of computations required grows a lot. A really nice thing to do is to row reduce the matrix to what is called an upper triangular (means all the entries below the main diagonal are zero). Then the determinant is simply the product of all the entries on the main diagonal. If you accessed this video from the Linear Algebra playlist, this method will be presented soon.
  Comment on newbarker's post “Yes, that would be a vali...”
  (18 votes)
a.somjp
Posted 4 years ago. Direct link to a.somjp's post “made me chuckle: "I don't...”
made me chuckle: "I don't want to beat a dead horse by showing you all of the different ways to find a determinant, but it might be useful to beat this dead horse"
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(6 votes)
Answer
sprangersw
Posted 8 years ago. Direct link to sprangersw's post “The outcome of this Det (...”
The outcome of this Det (A) = 45, while in the past video with the 3 x 3 matrix the outcome of the Det (A) was 35. How?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Phakawat Wangkriangkri
  Posted 7 years ago. Direct link to Phakawat Wangkriangkri 's post “a(3,1) in the past video ...”
  a(3,1) in the past video is 1, but here it's -1.
  Button navigates to signup page
  (4 votes)
Cole Elliott
Posted 8 years ago. Direct link to Cole Elliott's post “I am just learning the Ru...”
I am just learning the Rule of Sarrus in my algebra 2 class and was asked by my teacher to find out why this works. Is there a reason or is it just something that works and we all do it? Thanks!
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- InnocentRealist
  Posted 7 years ago. Direct link to InnocentRealist's post “Well, many of us don't kn...”
  Well, many of us don't know why using "minors and cofactors" works.
  
  Does your teacher want to find out the answer to his question, or was it meant just as an exercise for you?
  
  If you accept that minors and cofactors is accurate, then Sal has proved Sarrus' rule in the video based on that (I think Sarrus' rule is only for 3x3 determinants).
  
  If you want to prove Sarrus' rule from definitions, then here's the definition of a determinant:
  
  "def.: determinant: 2 (mathematics): A square array of numbers bordered on the left and right by a vertical line and having a value equal to the algebraic sum of all possible products where the number of factors in each product is the same as the number of rows or columns, each factor in a given product is taken from a different row and column, and the sign of a product is positive or negative depending upon whether the number of permutations necessary to place the indices representing each factor's position in its row or column in the order of the natural numbers is odd or even."
  
  For the first product, aei = a11a22a33, there are no permutations, so this product is positive.
  
  For bfg = a12a23a31, the second subscripts are in the order (2, 3, 1) (the first subscripts are always in numerical order (no permutations) to prevent confusion). To get from (1, 2, 3) to (2, 3, 1) first switch 2 and 1 to get (2, 1, 3) then switch 1 and 3 to get (2, 3, 1). that's an even number of permutations (two), so bfg = a12a23a31 is positive.
  
  Similarly cdh is positive.
  
  For ceg = a13a22a31, to get from (1, 2, 3) to (3, 2, 1), switch first 1 and 2, then 1 and 3, to get (2, 3, 1), then switch 2 and 3 to get (3, 2, 1); which is 3 permutations (an odd number), so it's "-ceg". Similarly you get "-afh" and "-bdi".
  
  So now we have proven that all the terms are correct. Now you can sketch the pattern (A augmented by repeating its first two columns, with ovals around all the resulting diagonals) of the elements as shown in the video, which helps you remember the terms and their signs. \\
  Comment on InnocentRealist's post “Well, many of us don't kn...”
  (2 votes)
arkorolls
Posted 10 years ago. Direct link to arkorolls's post “To find determinants, wou...”
To find determinants, wouldn't it be easier just to use expansion by minors?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Lucas Van Meter
  Posted 10 years ago. Direct link to Lucas Van Meter's post “Yes, it sure can be. If y...”
  Yes, it sure can be. If you have a lot of zeros in the matrix it is a great way to do it by hand.
  
  In general I believe that there are faster ways to compute the determinant by factoring the matrix into the product of a permutation matrix, a diagonal matrix, and an upper triangular matrix.
  Button navigates to signup page
  (2 votes)
45acegame
Posted 13 years ago. Direct link to 45acegame's post “Would this work for a fou...”
Would this work for a four by four or two by two? thanks
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(4 votes)
Answer
- Kyle Delaney
  Posted 9 years ago. Direct link to Kyle Delaney's post “You don't really need any...”
  You don't really need any special shortcuts for a 2x2, but you may notice that finding the determinant for a 2x2 matrix is already kind of similar to the Rule of Sarrus.
  Button navigates to signup page
  (2 votes)
ruec
Posted 10 years ago. Direct link to ruec's post “Does anyone know whether ...”
Does anyone know whether Khan's done any other videos or questions that ask you to solve a determinate by simplifying the determinate first i.e. taking out a common factor first then working out the determinate?
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(1 vote)
Answer
- InnocentRealist
  Posted 7 years ago. Direct link to InnocentRealist's post “Let's say there was a com...”
  Let's say there was a common factor of 2 (each element of the matrix is multiplied by 2). Then the products @
  2:00
  would be:
  
  8ae - 8afh - 8bdi + 8bfg + 8cdh - 8ceg. So det(2A) = 8*det(A), or det(2A) = 2^3(det(A) if A is 3x3.
  
  Similarly det(mA) = (m^n)det(A) if A is nxn.
  Button navigates to signup page
  (1 vote)
Anna
Posted 10 years ago. Direct link to Anna's post “why do they consider: 4x4...”
why do they consider:
4x4 determinant = 3x3 determinant + 3x3 determinant - 3x3 determinant = 2x2 + 2x2 - 2x2 + 2x2 - 2x2 + 2x2 - 2x2 + 2x2 - 2x2 recursive instead of iterative? I mean isn't 3x3 determinants iteration 1 and 2x2 determinants iteration 2?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- InnocentRealist
  Posted 7 years ago. Direct link to InnocentRealist's post “You need to define what y...”
  You need to define what you mean by iteration, and why you think expansion by minors and cofactors is iteration; because there aren't any successive values of the determinant - there is no answer until you get down to the 2x2 level. Recursion and iteration both entail repetition, but they aren't the same.
  
  Recursive: 2. a. Math. and Logic. Designating a repeated procedure such that the required result at each step is defined in terms of the results of previous steps according to a particular rule (the result of an initial step being specified).
  
  Iteration: b. Math. The repetition of an operation upon its product, as in finding the cube of a cube; esp. the repeated application of a formula devised to provide a closer approximation to the solution of a given equation when an approximate solution is substituted in the formula, so that a series of successively closer approximations may be obtained; a single application of such a formula; also, the formula itself.
  Button navigates to signup page
  (1 vote)
David O'Connor
Posted 8 years ago. Direct link to David O'Connor's post “If you refactor the expre...”
If you refactor the expression shown at
2:00
in a different way, you can see an algebraic demonstration of why the previous video (Determinants along other rows/cols) works.
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(1 vote)
Answer
- InnocentRealist
  Posted 7 years ago. Direct link to InnocentRealist's post “Cool. There are 6 differe...”
  Cool. There are 6 different ways to calculate the determinant (3 rows and 3 columns) by minors and cofactors. They all give the same 6 terms @
  2:00
  when you multiply them out.
  Button navigates to signup page
  (1 vote)

Video transcript

I don't want to beat a dead horse by showing you all of the different ways to find a determinant, but it might be useful to beat this dead horse because you'll see it done in different ways, in different context. And I thought I would at least show you that what we have been covering, so far, is very consistent with a way of determining, or finding determinants, that you might have been exposed to in your Algebra Two class. It's called the rule of Sarrus. Let me just prove it for you. Let's say we want to find a determinant. Let's say you want to find this determinant. So our matrix is a, b, c, d, e, f, g, h, i. We know how to do this. This is equal to, let's just go down that first row, a times the determinant of e, f, h, i minus b times the determinant of d, g, f, i plus c times the determinant of d, e, g, h. And what are these equal to? This is going to be equal to a, so let me write this, a times ei minus fh. And this is going to be minus b times dI minus fg. This is going to be plus c times dh minus eg and if we multiplied this out we get this is equal to aei minus afh minuses bdi plus right, minus times a minus, plus bfg plus cdh minus ceg. Now let me group the positive and the negative terms. So this term is positive, this term is positive and that term is positive. So we have this being equal to aei plus bfg plus cdh. Those are our positive terms. And then our negative terms are here. We have that term, that term, and that term. So we have minus afh minus bdi minus ceg. So this is a formula for the determinant of this matrix right here. Let's see what it actually looks like. Let me rewrite it. Let me rewrite our matrix. We do it in green. So we have a, b, c, d, e, f, g, h, i. We wanted to find its determinant. So let me show you something interesting here. aei is what? aei is a product of this guy, this guy, and that guy. So, you're essentially going along that diagonal right there. Now what is bfg? You're going this guy, this guy, and then you're going all the way down to this guy. So it's like if you imagine that when you come out of this side, you come out of this side. There's some video games where you go out one end and you end up showing up on the other end like that. It would also be a diagonal. Or even a better way to visualize it, let me redraw these two columns. Let me augment this determinant. It's not official terminology, but I think you'll get what I'm trying to do. So if I write these first two columns again. a, d, g, and b, e, h. This guy right here bfg, it's this one right here, this diagonal right there. And then you might guess what's about to happen. Where is cdh? It's this diagonal. It's that diagonal right there. So you take this product, add it to this product, add it to this product. And then you subtract these guys. Now what are these guys? Where is the afh? That one right there. So you subtract out afh, and then you subtract out bdi. bdi is that one right there. And then you have ceg, which is this one right there. So the Rule of Sarrus, sounds like something in The Lord of the Rings. The Rule of Sarrus is essentially a quick way of memorizing this little technique. You write the two columns again, you say, ok, this product plus this product plus this product, minus this product minus this product minus that product. Let's actually do it with the 3 by 3 matrix to make it clear that the Rule of Sarrus can be useful. So let's say we have the matrix, we want the determinant of the matrix, 1, 2, 4, 2, minus 1, 3, and then we have 4, 0, minus 1. We want to find that determinant. So by the Rule of Sarrus, we can rewrite these first two columns. So 1, 2, 2, minus 1, 4, 0. We rewrote those first two columns. And to figure out this determinant we take this guy. What is this going to be? 1 times minus 1 times minus 1. That is just a 1. Right, the minuses cancel out. Plus this guy, plus this product right here. I should draw a little bit neater. So what is this? 2 times 3 times 4. 2 two times 3 is 6. 6 times 4 is 24, plus 24. And then we take this guy right here. 4 times 2 times 0, anything times 0 is a 0. So that's going to be plus 0. And then we subtract out these guys. So you have 4 times 4, times minus 1. That's minus 16. It's minus 16, but we're going to be on the minus side of things. So it's 4 times minus 1 times 4, is minus 16. But since were going to do a minus on it, it's going to be plus 16. So it's 16. Then you have a 0 times 3 times 1. That of course is going to be 0. Would be a minus 0, but we can ignore it. So we can say plus 0 or minus 0 same thing. Then you have a minus 1 times 2 times 2. So that's 4 times minus 1 which is minus 4. When you go in this direction, from the top right to the bottom left, you are subtracting. So this would be a minus 4 but since we're subtracting, this becomes a plus 4. So the value of our determinant is equal to, by the Rule of Sarrus, we're going to have 16 plus 4 is a 20. 20 plus 25 which is equal to 45. So that actually is, I'd have to say, a faster way of computing this 3 by 3 derivative. And I just want to show you this is completely equivalent to the definition that I introduced you to a couple videos ago.