Algebra (all content)
Challenging complex numbers problem: complex determinant
Paper 1 Problem 53 from the challenging 2010 IIT JEE exam, about determinant of a complex numbers matrix. Created by Sal Khan.
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- At the end of the video, Sal says that the number of complex numbers that satisfy the solution of z is 1. The only solution is 0, and 0 is not a complex number, so why is there 1 complex number that satisfies the solution of z?(7 votes)
- Sal means "complex number" in the more general sense which includes pure real numbers (and pure imaginary numbers) as also being complex. Actually, unless someone is being really specific, they will always mean "complex number" in that sense from this point forward.(7 votes)
- When calculating the determinate why did he only use the first row?(3 votes)
- We can use any row or column for calculating the determinant. He just randomly used the first row.(3 votes)
- At10:34, how come w^4 isn't equal to w^2^2 (since w^6 is w^3^2)? 😮(2 votes)
- w^4 is equal to w^2^2 if you happen to square the w^2 's value you would get same value as the one calculated by Sal.(2 votes)
- I don't understand why he said the determinant should be equal to zero, and then calculate the determinant, rather than just solving this matrix as you do for an homogeneous system of equations = 0 ?(1 vote)
- Notice that the matrix has straight vertical bars on the side | | rather than the brackets [ ] that usually signify a matrix. The vertical lines tell us that this is a determinant.(3 votes)
- At12:50he says that z=0 is the only solution but the complex no. is 1.How?(1 vote)
- He says z=0 is the only solution for the equation and number of solutions for the equation is 1 since only z=0 satisfies the equation(3 votes)
- Was that a fast fourier transform matrix?(2 votes)
- Yes Sal, please make a video on this if you haven't already!(1 vote)
- I don't understand how Sal used the determinant chart. Can anyone help me?(1 vote)
- In a problem, when the value of a matrix is set to a constant, is it automatically assumed the problem is asking for the determinant of the matrix? I was a bit confused when Sal started finding the determinant of the matrix.(1 vote)
- how did you get the height to be sq (3)/2 and base to be -0.5??(1 vote)
- At1:07Why did you swap signs sir?(1 vote)
Let omega be the complex number cosine of 2 pi over 3 plus i sine of 2 pi over 3, then the number of distinct complex numbers z satisfying this determinant equaling 0. So we have this 3 by 3 determinant equaling 0. So let's just evaluate this determinant and see if we can solve for z, or figure out how many complex numbers z that we get satisfying this-- what's essentially an equation. So let's evaluate the determinant. So if we start with this term up here it's going to be this term times the determinant of this sub 2 by two matrix. So it's going to be z plus 1, times z plus omega squared, times z plus omega, minus 1. So this is the determinant of the sub 2 by 2 matrix right over here-- the sub 2 by 2 determinant-- and that's the z plus 1. And then we have to swap signs. We go here. We want to put a negative sign out front. You have that checkerboard pattern when you evaluate determinants. So minus omega times-- you block out that row, that column. So the sub-determinant is omega times z plus omega. Let me just multiply that out right now. So that's z omega plus omega squared, minus omega squared, times 1. So that's pretty neat. We already got something that simplifies a little bit. And then we have to worry about this omega squared. So it's going to be plus omega squared times omega times one-- omega times 1 is omega-- minus omega squared times z plus omega. So minus omega squared z minus omega to the fourth. And remember this whole expression needs to be equal to 0. This whole thing needs to be equal to 0. Now let's see if we can simplify this thing a little bit. So let me first multiply these two guys in here. So z times z is z squared. Z times omega is z omega. Omega squared times z is omega squared z. And then omega squared times omega is omega to the third power-- plus omega to the third power. And then we have this minus 1 here. And then all of that's going to be multiplied by z plus 1. Let me just continue with this blue part since we're already focused here. So this is going to be z times all of this. So it's going to be z to the third power plus z squared omega, plus omega squared z squared, plus omega to the third z minus z. I just multiplied z times all of this. And then plus 1 times all of this. So plus this thing again, z squared plus z omega, plus omega squared z, plus omega to the third, minus 1. And then let's simplify this over here. So that in green, these canceled out. So we're just left with-- I'll do the same green-- negative omega times z omega. So it's minus z omega squared. And then over here in magenta we have plus omega to the third power minus omega to the third z. Oh, let me be careful. That would be omega to the fourth. Omega squared times omega squared is omega to the fourth-- omega to the fourth z. And then omega squared times omega to the fourth is omega to the sixth. And we have a negative sign. So negative omega to the sixth. And of course, this whole thing needs to be equal to 0. Well, let's see if we can simplify this. Let's just group the terms-- the different powers of z. So we have a z to the third here. And that is the only z to the third that we have. And then let's group the z squared terms. So this is a z squared term. This is a z squared term. That's a z squared term. And then we don't have any more z squared terms. And so the coefficients-- this is going to be the same thing as omega squared-- that's this one right over here-- plus omega, plus 1, times z squared. So we've taken care of everything that I've underlined in pink over here. Now let's worry about the z terms. I'll do it in this color. So this is a z term. We're just multiplying times z. This is a z term. This is a z term. That is also a z term. And then do we have any other z terms? Well, this is a z term right over here. And in fact, these two cancel out. You have negative z omega squared, positive z omega squared. So this and this cancel out. And so our only z terms are these three over here. So we have plus omega to the third. And then we have plus omega to the third. Oh let me-- this color is too close. Plus omega to the third. And then our next z is minus 1. But then we actually have this omega here. So plus omega minus 1. I was getting confused because this white is so close to this pink. So it's omega to the third minus 1, plus omega, times z. And then oh, we can't forget this. We have this term over here. Omega to the fourth times z. So we have minus 1 omega to the third z. Then we have minus 1 z. Then we have plus omega z. And then we have negative omega to the fourth z. So I'll just put it out here. I'm not doing it in a nice descending power order, but I don't want to have to rewrite all of this. So then we took care of all of the z terms. And then what we have left is just what we can view-- the constant terms, from a z point of view. So we have omega-- So plus omega cubed. Actually, we have two omega cubed. We have an omega cubed here and an omega cubed there. We've taken care of those. And then we have minus omega to the sixth. And then we have minus 1. And all of this needs to equal 0. Now, this might look pretty daunting, especially when omega looks like this fairly complex-- well, it is a complex expression. But maybe, or probably, it popped into your mind that this can also be expressed as an exponential using Euler's formula. We know that e to the i theta is equal to cosine of theta plus i sine of theta. So omega can be rewritten as, omega is the exact same thing as e to the 2 pi over 3 i. That's what omega is. And what is e to the 2 pi over 3 i? We can evaluate that here. It's cosine 2 pi over 3. And we can get our unit circle out. 2 pi over 3 corresponds to 120 degrees. So let me get my unit circle out just so we can figure out what the values are. And the reason why I'm doing this-- I'm actually going to evaluate it here just to get a number here-- but the reason I'm doing this is it becomes much easier to take the powers of e to the 2 pi over 3 i than to take the powers of this thing over here. But if we take the unit circle, if we're at 120 degrees on the unit circle like this, this is 2 pi over 3. Then we have a situation. This angle over here is going to be 60 degrees. And so our height is going to be square root of 3 over 2. And our x-coordinate is going to be negative 1/2. So this is going to be equal to negative 1/2. Cosine of 2 pi over 3 is negative 1/2. And then our sine of 2 pi over 3 is square root of 3 over 2. But that's being multiplied by an i. So plus square root of 3 over 2 i. So that is omega. Now, let's think about omega squared. Omega squared is going to be this raised to the second power. So it's e to the-- we just multiply it by 2. e to the 4 pi over 3 i. We just multiply this by 2. So it's equivalent to 240 degrees. So it's equivalent to this right over here. So let me write it. This is the same thing as cosine of 4 pi over 3, plus i sine of four pi over 3. And here our x-coordinate is going to be the same. It's going to be negative 1/2. So this is equal to negative 1/2. And our y-coordinate is negative square root of 3 over 2 i. Now let's think of omega to the third power. Omega to the third power is equal to e to the 2 pi i, which is equal to cosine of 2 pi plus i sine of 2 pi. Well, sine of 2 pi is just 0. Cosine of 2 pi is equal to 1. Then we go to omega to the fourth power, because that also shows up in these expressions. So we have omega to the fourth power is equal to-- we just multiply it times 4-- it's e to the 8 pi over 3 i. 8 pi over 3 is equivalent to-- so this was 2 pi over 3. This is 4 pi over 3. This is essentially 6 pi over 3. And then 8 pi over 3 becomes-- is essentially the same angle as 2 pi over 3-- so its value is going to be the same. Negative 1/2 plus square root of 3 over 2 i. And then we don't have an omega to the fifth, but we do have an omega to the sixth here. So let's just figure that out. Omega to the sixth-- well, that's just omega cubed squared. And omega cubed is 1. So it's just going to be 1 squared, or it's also going to be 1. Now with that out of the way, let's see what these things simplify to. Hopefully they simplify to something. So this expression right here-- omega squared-- we have-- I'll write it up here. We have negative 1/2 minus square root of 3 over 2 i. That's omega squared. Then we have plus omega. Omega is right here. Negative 1/2 plus square root of 3 over 2 i. And then we have a plus 1. So the negative square root of 3 over 2 i and the positive square root of 3 over 2 i cancel out. Negative 1/2 minus 1/2 is negative one, plus one. Those cancel out. So this whole expression comes out to be 0. So we can ignore. This is going to be 0 no matter what. Now let's look at this over here. This is omega cubed. Omega cubed we know is 1, plus omega, which is-- so I should write minus 1/2 plus square root of 3 over 2 i, minus 1, minus omega to the fourth. Well, omega to the fourth is the same thing as omega. Omega to the fourth is the same thing as omega. So you have omega minus something that's the same thing as itself. Those are going to cancel out. Omega cubed is 1. So the 1 and the 1 are going to cancel out. So this is also equal to 0. I think I see a pattern here. And then finally, let's look at these characters over here. 2 times omega cubed-- this is just going to be a 2 here-- minus omega to the sixth. Well, that's just minus 1. And then you have a minus 1. So this 2, minus 1, minus 1. This is also equal to 0. So that whole determinant that whole equation has simplified to z to the third power is equal to 0. And the only number, that when they take it to the third power-- real, or complex, or anything-- is going to be 0. z equals 0 is the only solution. But they're not asking us that. They're asking the number of distinct complex numbers z satisfying this. So z equals 0 is the only one that satisfies this. So the number of complex numbers satisfying it is 1.