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Determinants along other rows/cols

Finding the determinant by going along other rows or columns. Created by Sal Khan.

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  • spunky sam blue style avatar for user Ethan Dlugie
    Is it possible to find the determinant by going down the diagonal?
    (10 votes)
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    • piceratops ultimate style avatar for user Paul
      Yes, and no. One method of finding the determinant of an nXn matrix is to reduce it to row echelon form. It should be in triangular form with non-zeros on the main diagonal and zeros below the diagonal, such that it looks like:
      [1 3 5 6]
      [0 2 6 1]
      [0 0 3 9]
      [0 0 0 3] pretend those row vectors are combined to create a 4x4 matrix. Once it is in that form so that it appears like:
      1 3 5 6
      0 2 6 1
      0 0 3 9
      0 0 0 3
      Then the determinant = the product of the entries along the diagonal, such that determinant = (1)(2)(3)(3) = 18.
      Note* if the main diagonal contains a zero the determinant is also 0, thus the matrix is not invertible.
      Hope that was clear enough to help.
      (19 votes)
  • leafers tree style avatar for user Vilhelm Urpi Hedbjörk
    Am I the only one that doesn't see why it works going down any row or column, and why you need to switch signs?
    (5 votes)
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    • aqualine ultimate style avatar for user Kyler Kathan
      Let's start with a 3x3 determinant.
      |a₁ a₂ a₃|
      |b₁ b₂ b₃|
      |c₁ c₂ c₃|
      If we multiply it out, we get:
      a₁b₂c₃ - a₁b₃c₂ + a₂b₃c₁ - a₂b₁c₃ + a₃b₁c₂ - a₃b₂c₁
      Notice how each term has an a, b, and c in it and also has a 1, 2, and 3 in it. This is why it works to use any row or column. Whichever row or column you use is the one you're factoring out. So, let's say we want to use the 2 column. In doing so, we factor out all of the 2's:
      a₂(b₃c₁-b₁c₃) + b₂(a₁c₃-a₃c₁) + c₂(a₃b₁-a₁b₃)
      Notice that in each of the parenthesis, we have the equation of a 2x2 determinant now. However, 2 of them go 31-13 while the other goes 13-31. If we want it to be the determinant of a sub-matrix, we need them to be in the order 13-31, so we get:
      -a₂(b₁c₃-b₃c₁) + b₂(a₁c₃-a₃c₁) - c₂(a₁b₃-a₃b₁)
      This is why it switches signs depending on which column or row you choose.
      (6 votes)
  • old spice man green style avatar for user Mr. Lunetta
    Time stamp not showing for me, but at about 25% Sal give the formula Sign(i, j) = -1^(i+j). That should be written (-1)^(i+j) so the negative will also be raised to the power.
    (5 votes)
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  • starky ultimate style avatar for user Thales Alexandre
    If having 0s make finding the determinant easier, is it better to reduce it to rref first and then take the dets?
    (3 votes)
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  • duskpin ultimate style avatar for user Mayesha Fairuz
    if the entries are variables instead of number should we use the same process??
    (2 votes)
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  • blobby green style avatar for user Katey Strader
    Can you find this on a graphing calculator?
    (1 vote)
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  • blobby green style avatar for user attero.dx
    So the row or column that you choose to perform the cofactor expansions will be your pivot row to perform row or column operations from? Or can you switch your pivot row or column at any time during the process? Will that effect the process of computing the determinant?
    (1 vote)
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  • blobby green style avatar for user r4kd.Martin
    For this method, could we also use coloumn or row reduction to for example, get a row of [1 2 0 0] into [1 0 0 0] by doing C2 - 2C1 = C2 and then go about your method?
    (1 vote)
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  • blobby green style avatar for user Mez Cooper
    The videos in this section are beautiful. Excellent this one too, I didn't know you could keep changing where you did the determinant. I usually chose one row for the overall determinant and all the "inner" determinants on the same row.
    (1 vote)
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  • blobby green style avatar for user Andrea Andrews
    Is this process also considered finding the M32 and C32 of a 4 x 4 matrix?
    (0 votes)
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Video transcript

In the last video, we evaluated this 4 by 4 determinant and we found out that it was equal to 7. And the way we did it is we went down this first row. We used the definition I gave you in the last few where use this first row. I could even write it here. We said this is equal to 1 times the determinant of 0, 2, 0. 1, 2, 3. 3, 0, 0. Minus 2 times the determinant. You cross these guys. Cross that row and column out. 1, 0, 2. 1, 0, 2. 2, 2, 0. 0, 3, 0. And then we went to the plus the 3 times its sub matrix. I don't have to figure that out. Just cross out that row and that column. And then minus 4-- just keep switching the sign-- times the determinant of its sub matrix. So this one had a bunch of terms. And this one's going to have a bunch of terms. Cross that row and that column out. You get 1, 0, 2. 0, 1, 2. 2, 3, 0. I'll just write it here. 1, 0, 2. 0, 1, 2. 2, 3, 0. And that is a completely legitimate way to figure out a determinant. And that was our definition for how to find a determinant. But I want to show you in this video that there's more than one way to solve for determinant. What I'm going to show you this way is the same thing that we did down this first row. We can actually do down any row or any column of this determinant, or of this matrix. And the reason why that's useful is because we can pick rows or columns that have a unusually large number of zeros. Because that tends to simplify our computations. So the first thing that you have to do before you embark on picking an arbitrary row or a column, let's say, for example, we want to pick this. Let's do one row and one column in this example. So let's say we want to go down that row instead because we like the fact that has a lot of zeros there. The first thing you have to do is remember the pattern. Remember, you switch signs on the coefficients. You don't just switch signs as you go down a row. You also switch signs as you go down a column. So the general pattern for 4 by 4 will look like this. It'll be plus, minus, plus, minus, minus, plus, minus, plus. And then you get a plus, minus, minus, plus, plus, minus, minus, plus. It's really this checker board pattern. If you wanted to figure out the sign for any ij. So let's say you wanted to figure out the sign for, this is 2, 2. So if you want to find the sign-- let me write it this way. Say we have a function, let me define a function. I think the checker board pattern's pretty clear to you, but I'll just write it down. Let's say I wanted to figure out the sign-- the sign not the trig ratio. I want to figure out the sign of any entry where you give me an i and a j. What you could do is you just take negative 1 to the i plus j power. So if you wanted to figure out the sign for the, let's say, you are in row 4, column 2. Row 4, column 2. So what will it be? Do 4 plus 2. Negative 1 to the sixth power is equal to 1. That's going to be a positive 1. Let's say you want to take this guy. So this is i is equal to 2. j is equal to 3. We're in the second row, third column. 2 plus 3 is 5. Minus 1 to the fifth power is minus 1. And you have a minus there. So that's another way to think about it. But the checker board pattern's pretty straightforward. So now that you have the checker board pattern in your mind, let's go down this row. Let's go down this row. So we start with the 2. But notice that we have to multiply by minus. Because we go plus, minus, plus, minus. So you have a minus 2 times the determinant of its sub matrix. So you cross out this row and that column. And you're left with this matrix up here. So it's 2, 3, 4. 0, 2, 0. 1, 2, 3. This is a minus and then you have a plus. So you have a plus 3 times-- get rid of this guy's column and row. And you have 1, 1, 0. 1, 1, 0. That's that right there. 3, 2, 2. 3, 2, 2. And you have 4, 0, 3. 4, 0, 3. And then you would have a minus a 0 times its sub matrix plus a 0. But we can ignore those because 0 times anything is a 0. So already we've simplified our determinant a good bit. So let's see if we can evaluate this and get the same number. Because only then will it be reasonably satisfying. So what's the determinant of this guy? Well we can do the exact same principle. We can go down any row or column that seems to be especially simple. So let's go down that row. Because that row seems especially simple. So this is going to be minus 2. That's this minus 2 right here times the determinant of this guy. So the determinant of this guy, we just have to go and say OK we have a plus, we have a minus, and then we have a plus. It's going to be minus 0 times its sub determinant, I guess we would call it. Get rid of that row or that column and that row. So it'd be minus 0 times anything, that's just going to be 0 plus 2. So plus 2 times the determinant. Get rid of its row and its columns. 2, 4, 1, 2. 2, 4, 1, 3. And then you have a minus 0 times this thing. But who cares what that is because you have a 0 times. So that just simplified to that which is nice. Let me write it like that. And then you have plus 3 times this thing right here. We don't want to do the first row. We have no non 0 terms here. Let's at least do this row right here for a little bit of variety. None of the columns actually seem that interesting. They all have, at most, one 0. So if we do that one right here, this is a plus, minus, plus, minus, plus. So we'll have a plus 0 times 3, 4, 2, 0. We can ignore that. Minus 2 times the determinant. Get rid of this column, that row. I have to be very careful. I put this minus there, but there wasn't a minus 1 there. Let me write it like-- I want to make sure I don't make any careless errors right here. This is a plus 1. I just drew a minus 1 there to show you how things switch signs. So this is going to be a 3 times. So we're going to go down-- This is this 3 right here. I lost my bearings with that minus there. We're trying to find the determinant of this. So it's 0 times this matrix. We can ignore that. Minus 2 times its sub matrix. Which is that and that. So it's 1, 1, 4, 0. And then you have a plus 3 times its sub matrix. 1, 3, 1, 2. 1, 3, 1, 2. Just like that. Let's see if we can simplify this. 2 times 3 is 6 minus 1 times 4. So, this becomes 6 minus 4. So that's 2. So this whole thing simplifies to 2 times 2 which is 4. 4 times minus 2. The whole thing simplifies to minus 8. Now, this guy right here, we have 1 times 0, which is 0, minus 1 times 4. This is minus 4 times minus 2. This whole thing becomes a positive 8. You have 1 times 2, which is 2, minus 1 times 3. Which is minus 3. So you get a minus 1. So this becomes a minus-- I mean you get 2 minus 3 minus 1 times 3. So this becomes a minus 3. So you have an 8 minus a 3. So this becomes a minus five. You have a 3 times a minus 5 right there. So 3 times minus 5 is going to be equal to minus 15. Let me make sure I got-- oh I made a silly mistake. If you have an 8 minus a 3, this is 8, this is minus, this is going to be 5. Very easy. Your brain starts to get fried if you do this long enough. And the you have a 3 times a 5 and you get a 15. You get a 15. And then you have a 15. This term plus this term is 15 minus 8 which is equal to 7. Which, lucky for us, I barely evaded making a careless mistake there, we got the right answer. But this is a much simpler computation than we did in the last video. And it was much simpler because we picked the row that happened to have a lot of zeros on it. So we only had two of these terms instead of four terms like we had in the last video. And you could do the same, you could pick columns that have a lot of zeros. You just have to make sure that you always use this checker board pattern.