- Determinant when row multiplied by scalar
- (correction) scalar multiplication of row
- Determinant when row is added
- Duplicate row determinant
- Determinant after row operations
- Upper triangular determinant
- Simpler 4x4 determinant
- Determinant and area of a parallelogram
- Determinant as scaling factor
Determinant when row multiplied by scalar
The determinant when a row is multiplied by a scalar. Created by Sal Khan.
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- At5:31Why is the cofactor determinant of f equal to (dh-eg) and not (ah-bg)? I might have misunderstood the pattern.(57 votes)
- You are correct; Sal made a mistake there.(43 votes)
- Do the same rules for the determinant when a row is multiplied by a scalar apply to when a column is multiplied by a scalar?(8 votes)
- Yes. If you transpose a matrix its determinant doesn't change so you can consider multiplying a column by a scalar as first transposing the matrix, then multiplying the equivalent row by the scalar.(7 votes)
- Do these rules apply for a m by n matrix?(4 votes)
- The determinant isn't defined for non-square matrices because non-square matrices aren't invertible in the traditional sense.(13 votes)
- at ~13:20, sal is writing out 3rd/final term of the 3x3 determinant |A|...but shouldn't it be "-f(ah-bg)" vs "-f(dh-eg)" ?(5 votes)
- Yes. Transcription error from a few minutes in. [I'm writing words to pass the comment screener.](1 vote)
- Can anyone tell me why in 3*3 matrix determinant is of order + - + -...? What is its proof? I am talking about the cofactor. Where does the cofactor emerge from?(2 votes)
- How would we use the Anxn formula in finding the determinant for example a 4x4 matrice? Can somebody tell me what values to place in i and j? And how will we find the determinant of Aij in this case?(2 votes)
- How to multiply 4m+ 3n = 1 by 8/x - 9/y = 7 ?(2 votes)
If we make that substitution we get the two equations 4m+3n=1 and 8m-9n=7. Lets solve it by solving the first equation for m to get m=(1-3n)/4 and then substitution that into the second to get 8((1-3n)/4)-9n=7. This simplifies to 2-15n=7. Solving for n gives us n=-1/3. Then we remember we have m=(1-3n)/4=1/2.(1 vote)
- At5:59, shouldn't he have written
| a b |
| g h |?(2 votes)
- is there any general formula ?(1 vote)
- Sal's method of "minors and cofactors" is I guess a general formula. Based on two previous videos,
https://www.khanacademy.org/math/linear-algebra/matrix-transformations/inverse-of-matrices/v/linear-algebra-nxn-determinant, and https://www.khanacademy.org/math/linear-algebra/matrix-transformations/inverse-of-matrices/v/linear-algebra-determinants-along-other-rows-cols,
the (recursive) general formula for expanding along the ith row or jth column using minors and cofactors is:
|A| (expanding along row i) = sum(j=1,n) [(-1)^(i+j)*a_ij*A_ij], or
|A| (expanding along column j) = sum(i=1,n) [(-1)^(i+j)*a_ij*A_ij].
This gives you n products of elements ("a_ij") of A and their minors = (n-1)x(n-1) determinants "A_ij" (see the first video above) which can be solved by the same equations above and each yield n-1 products of elements and (n-2)x(n-2) minors, and so on to total n-2 recursions, until the minors are 2x2 and you get an answer.
There is another general formula using the definition of a determinant, which takes more time, and also explains more about what a determinant actually is and why "minors and cofactors" works.(2 votes)
- What does j mean? Is it he column?(1 vote)
- j is a column in A just like i is a row in A, so yes.(1 vote)
Let's explore what happens to determinants when you multiply them by a scalar. So let's say we wanted to find the determinant of this matrix, of a, b, c, d. By definition the determinant here is going to be equal to a times d minus b times c, or c times b, either way. ad minus bc. That's the determinant right there. Now what if we were to multiply one of these rows by a scalar. Let's say we multiply it by k. So we have the situation a, b. Let's multiply the second row times k. So we kc and kd. Now what's the determinant going to look like? We're going to have a times kd, or we could just write that as kad, minus kc times b, or we could right that as kbc. If we factor out the k, we get that equals k times ad minus bc. Now you immediately see that this thing is the same thing as this thing. So this is equal to k times the determinant of a, b, c, d. So when you just multiply a row by a scalar-- just one row, not the entire matrix-- when you just multiply a row by some scalar, the resulting determinant will be the original determinant times that scalar. Now you might say well what happens if I multiply the whole matrix times that scalar. Well that's equivalent to multiplying by a scalar twice. Let's say I have the matrix A, and the matrix A is equal to a, b, c, d. If I were to think about the matrix kA, now I'm not just multiplying one row. I'm multiplying the whole matrix by a scalar. This is going to be equal to ka, kb, kc, and kd. When you figure out its determinant, the determinant of k times A is going to be equal to the determinant of ka, kb, kc, and kd. Here you're immediately going to end up with k squared terms. You're going to have k squared times ad. It's going to be equal to k squared ad minus k squared bc, or k squared times ad minus bc, or k squared times the determinant of just A. So you have to be very careful. This is only for a two by two case. You'll find out if this was an n by n matrix that this would have been k to the n. So the take away is the only way you can say that this is going to be some scalar multiple times your original determinant, is only if you multiply one row times that scalar multiple, not the whole matrix. Let's see how this extends to maybe a 3 by 3 case. You might say, hey Sal, you just picked the second row. Does it work with the first row? I'll leave that for you to determine, but it does. It does work. It doesn't matter which row I multiplied it by. Let's take the three by three case. Let's say we have some matrix. Let's call this A again. I'm redefining A. It's going to be a, b, c, d, e, f, g, h, i. Then if you take its determinant, the determinant of A is going to be equal to-- we can do it a couple of different ways. But I'll just pick some arbitrary row. Because that's the row that we're going to multiply by some scalar. So let's just take that row right there. Remember the plus minus pattern. Remember plus, minus, plus, minus, plus, minus, plus, minus, plus, that little checkerboard pattern. So d is a minus right there. So it's going to be equal to minus d times the determinant of its submatrix. So you cross out that column in that row. It's b, c, h, i. It's going to be plus e times its submatrix a, c, g, i. It's going to be minus f times-- you get rid of that row in that column-- d, e, g, h, the determinant of d, e, g, h. That's the determinant of this matrix A. Now what if we define some new matrix here? Let's call it A prime. Let me make this scroll down a little bit. Let me define A prime right here. A prime, I'm just going to multiply this row by a scalar. So it's going to be equal to a, b, c, kd, ke, and kf. I'm not multiplying the whole matrix times the scalar. I can't say this is kA. I'm just multiplying one of its rows. Then I have g, h, and i. So what's the determinant of A prime going to be? I put that prime there. So it's different than A, but it's derived from A. I just multiplied one row of A times the scalar. Well I can go along that same row that I did up here. I can go along that same row. The only difference is is that instead of having a d, I now have a kd. Instead of an e, I now have a ke. So instead of a d, I'm going to have a kd there. Instead of an e, I'm going to have an ke there. So it's going to be this exact same thing, but I can replace this guy, this guy, and this guy with them multiplied by k. So it's going to be equal to minus kd times the determinant of the submatrix b, c, h, i-- I'm not even look over here because it's going to be the same thing as that one up there-- plus ke times the determinant of a, c, g, i, minus kf times the determinant of d, e, g, h. What is this equal to? This is equal to, if you just factor out the k, it's equal to k times this. So it's equal to k times the determinant of A. So our result also worked for three by three case. I just happened to pick the middle row. But I encourage you to pick other rows and to see what happens. So let's actually do it for the general case. Because I've just been giving you particular examples, and I like to show you the general proof when the general proof isn't too harry. So let's say I have an n by n matrix. Let's say that I have matrix A. Let's say that A is n by n. You can write it like this. This is the first row. This is the first column a11, a12, all the way to a1n. I'm going to pick some arbitrary row here that I'm going to end up multiplying by a scalar. So we can go down here. Let's say row ai, so this is ai1, ai2, all the way to ain. This is some row that I'm going to use to determine the determinant. Remember we can go to any row to get the determinant. Then finally you keep going. You get an1, an2, all the way to ann. This is as general as you can get for an n by n matrix. Now let's figure out its determinant, so the determinant of A. I'm just going to go down this row right there. So the determinant of A is equal to what? Well we have to remember the checkerboard pattern. We don't know where we are on the checkerboard pattern because I just picked an arbitrary general row here. But we can use the general formula that the sign is going to be determined by negative 1 to the i-- I don't know if i is even or odd-- so it's going to be i plus for this term 1 power. That's it's sign. This is what gives us the checkerboard pattern. Let me make that clear. It looks complicated, but this is just a checkerboard. Times this term right there, so times ai, so the coefficient, ai1, and then times this guy's submatrix. You remember the submatrix. You get rid of this row, and this column is going to be everything that's left over. So times that submatrix of ai1. Then it's going to be plus-- we just keep doing it-- plus negative 1 to the i plus 2 times ai2 times its submatrix-- all the way, you just keep going-- plus minus 1 to the i plus n times ai. You're in the nth column, and then its submatrix. This is going to be an n minus 1 by n minus 1 matrix. All of these are going to be. Just like that, that's the determinant of A. We could actually rewrite it in sigma notation. That'll simplify things a little bit. So the determinant of A, we can rewrite it as the sum from j is equal to 1 to j-- I'll write it explicitly here-- j is equal to n of negative 1 to the i plus j times aij, and then each of the submatrices, Aij. This thing right here is just another way of writing this thing I wrote up there. I'm just saying the sum. You just take j equal 1, put them in there. You get this term right there. You take j equal 2, you add it. You get this term right there. You keep doing it. You get j equal n. You get that term right there. So these two things are equivalent. So what happens I have some new matrix? Let me copy and paste my current matrix. So let me copy and paste it. Actually let me copy and paste everything. That'll make things move quickly. I copied it. Now let me paste it just like that. Let me define a new matrix A prime. It's still an n by n matrix. But that row that I just happened to use to determine my determinant, I'm going to multiply it by a scalar k. So it's k ai1, k ai2 k ain just like that. So what's the determinate of A prime? Well we're just going to go down this row again. But now instead of just an ai1, we have a k ai1 Instead of an ai2, we have a k ai2 Instead of an ain we have a k ain. So it's determinant is just going to be this same thing, but instead of an aij everywhere, we're going to have a k aij. So this is the determinant of A prime. We could just take out this constant right here. It has no i's or j's in it. I has no j's in it in particular, so we can just take it out. So it's equal to k times the sum from j is equal to 1 to j is equal to n of minus 1 to the i plus j times aij. This is the coefficient. This is the submatrix for each of those coefficients, aij. That's a matrix right there, an n minus 1 by n minus 1 matrix. Then you immediately recognize-- I think you saw where this was going-- this right here is just the determinant of A. So we get the result that the determinant of A prime is equal to k times the determinant of A. So we've just shown you in general, if you have any n by n matrix, if you multiply only one row, not the whole matrix, only one row by some scalar multiple k, the resulting determinant will be your original determinant times k. Now I touched on this in the original video. What is the determinant of k times A? So now we're multiplying every row times k. Or another way to think about is you're multiplying n rows times k. So you're doing this n times. So if you multiply k times itself n times, what do you get? You get k to the n. So this is going to be equal to k to the n times the determinant of A. If you just do it once, you get k times the determinant of A. Now if you do a second row, you're going to get k times k times the determinant of A. If you do a third row, you're going to get k to the third times the determinant of A. The fourth row, k to the fourth times the determinant of A. If you do them all, all n rows, you're going to have k to the n times the determinant of A. Anyway, hopefully you found that interesting. I encourage you to experiment with these other ways. Try going down a column and seeing what happens.