You see nonlinear equations, they see linear algebra! (Harvard-MIT math tournament) 

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i have another way xy + z + xz + y = 40 + 51 x(y+z) + y + z = 91 (x+1)(y+z) = 91, we know that y+z = 19 - x, so : (x+1)(19-x) = 91 19x - x^2 + 19 - x = 91 x^2 - 18x + 72 = 0 (x-12)(x-6) = 0 x = 12 or 6 then just plug x into the equation and we will get the same value of y and z like in the video, CMIIW :3

As they say, “linear algebra is limited to linearity property, but oftentimes, it is easy to fit your nonlinear problem in a linear form”

Add 1st and second equation (y+z)(x+1)=91 x+1+y+z=20 Take x+1 as a and y+z as b Then ab=91 and a+b=20 So (a,b)=(13,7),(7,13) x is either 6 or 12

I started Gauss-Jordan elimination on the matrix at 5:18, got to [1 1 19-x] [0 x-1 x+32] [0 1-x x²-19x+40] and concluded that -(x+32) = x²-19+40 (to make row 2 + row 3 = 0), which gives x=6 and x=12.

Adding first two gives (x+1)(y+z) = 91 Final gives (x+1)+(y+z) = 20 Solving for x+1 and y+z gives the pair (7, 13), in either order. Backsubstituting in the first eqns, et voila.

I have another way x + y + z = 19 xz + y = 51 xy + z = 40 y = 51 - xz z = 40 - xy x + 51 - xz + 40 - xy = 19 x - xz - xy = -72 -x (y + z - 1) = -72 x (y + z - 1) = 72 x + y + z = 19 then y + z - 1 = 18 - x x (18 - x) = 72 18x - x² - 72 = 0 x² - 18 x + 72 = 0 (x - 6) (x - 12) = 0 x1 = 6 x2 = 12 Next: y + z = 13 y + 6z = 51 6y + z = 40 6y + z = 40 3y + 3z = 39 3y + 18z = 153 (Substracting) -20z = -152 z1 = 7,6 y1 = 5,4 y + z = 7 y + 12z = 51 12y + z = 40 12y + z = 40 6y + 6z = 42 6y + 72 z = 306 (Substracting) -77z = -308 z2 = 4 y2 = 3 Solutions: x1 = 6 y1 = 5,4 z1 = 7,6 x2 = 12 y2 = 3 z2 = 4

I combined the first two equations, did some algebra to get (x+1)(y+z)=91, used the third to replace (y+z) with (19-x), and arrived at that same quadratic. After which I promptly made a sequence of arithmetic errors solving the quadratic and then subsequently even more arithmetic errors solving the resulting system of equations. It turns out that if you fail at basic arithmetic, math is hard. :)

I love how the math you do is simple and understandable, but used creatively.

Love the way u solve it. Thx. ❤

2:43 I love how these 3 equations actually do still have a valid real number solution

I was doing this method of eigenvalues to solve systems of differential equations at my differential equations class today. Good lecture professor!

2:32 was is a coincidence because 5y+2z=10,y+z=3 and 2y-z=1 has a solution which is (y,z)=(4/3,5/3)

6:16 determinant of a determinant ?

Proud to have been able to solve this problem

Loved this approach, thinking out of the box

More of these please 🙏🙏

Adding the first two equations, you get (y + z)(1 + x) = 91. From the second equation: y + z = 19 - x. Substitute and get (19 - x)(1 + x) = 91. Solving for x with Bhaskara's Formula, you obtain x = 12 or x = 6, and so on.

Your digression about the 0 determinant not always giving solutions also applies in reverse. If you look at where the first two equations are linearly dependant, that must give 0 for the determinant, which gives you a root of the cubic (x=1) without needing to notice anything about the coefficients.

In the example at 2:30, the 3rd equation is actually the 1st minus 3 times the 2nd, meaning it is also redundant

Linear algebra is one thevmost useful math tools i never used unless forced to in my undergrad

Time for cheap tricks; Computing L1+L2-L3- x L3 on the left hand side... we have (xy+z)+(xz+y)-(x+y+z)-x(x+y+z) carefully cancelling...we are left with -x-x² The right hand side is 40+51-19-19x=72-19x -x-x²=72-19x Then I got lazy... but we arrive at the same polynomial of x(up to a non-zero scalar multiple) so it should arrive at the same solution

Add the first 2 equations, factor and take 2 cases, you are done

You don't need to solve the cubic, you can just say that it's easy to see that the first two rows of the 3x3 system of equations are linearly independent, therefore the third row must be expressible as a linear combination of the first two. Then looking at the left hand side, the respective coefficients must be 1/(1+x) and 1/(1+x), which yields 91/(1+x) = 19-x, x=6 or x=12.

What happens if you equate phytagoras Therom to the quadratic formula or is that not possible

I haven't heard about any of those things but it sounds cool.

Beautyfull. Thank you very much because you helped me to solver this exercise of Lineal Algebra.

Another way is let w = y + z and it can form an quadratic equation

2:42 y = 4/3, z = 5/3. The third equation works just fine....

it is actually solvable using gaussian elimination 0 x 1 | 40 0 1 x | 51 1 1 1 | 19 replace r1 and r3 r3-xr2 r1-r2 r2-xr3 r1-(1-x)r3 doing these steps gives and identity matrix and gives us x, y and z with respect to x x=(51x-40)/x+1 -32 then we get a quadratic equation that gives x=6 or 12 y=51- 51x^2-40x/x^2-1 z=(40-51x)/(1-x^2) we can then substitute 6 and 12 into these and get y and z s1 (6, 5.4, 7.6) s2(12, 3 ,4)

How much would you bet that whoever came up with this problem started with the functions x*y+z, x*z+y, and x+y+z, picked some values for x, y, and z, and plugged them in to get the values x*y + z = 40, x*z + y = 51, and x+y+z = 19, only to be surprised later when it had multiple solutions?

Can you make a video on how to use Weierstrass factorization theorem?

Any closed contour integral videos?

Hey I am an high schooler, I didn’t know most of the determinant formulas you used , but your video was fascinating. Thank you.

hi, thanks for the video, it's interesting as always, do you have a video where you detail more the method you're using at 10:00 for finding the factors of the cubic equation ? also I tend to have the issue in maths that I try to brute-force a lot of calculations without thinking to use methods that would be easier, like in this case using substitutions and big annoying calculations instead of just thinking to use the determinant like you did, so do you have any method beyond just habit, training and exercises to have some idea of what methods to use whenever you see a new math problem ? is there anywhere you might check online if you're not sure what to do ? I'd love to see how your thought process goes from the moment you discover the problem

Suming the first two equations gives (x+1)(y+z) = 91 and the third equation gives x+1+y+z=20 thus x+1 = 10 +/- sqrt(100-91) = {13,7}. In the first case (x=12) we get (Cramer) y=(40x-51)/143=429/143=3 and z=(51-3)/12=4. In the second case (x=6) we get (Cramer) y=(6×40-51)/35=189/35=27/5 so z=40-27*6/5=38/5. So, after checking they work, the set of solutions is {(12,3,4),(6,27/5,38/5)}. EZ but tedious.

THANK YOU for this idea, can't believe is was that simple to fix x as a constant and take determinants - since x is non-zero! Ignore all of the comments talking about the trivial way to solve this with gauss-elim, they don't understand the power this method has

Yess I asked for more last video and got more 😂

wow, i like how you using matrix as part of solving

The way I had solved this I got values which when added and multiplied givens answers close to 40,51,19. The set of answers was (x,y,z)=(7.3,4.6,6.2) and this is the only set I got.

Got 'em both!

Just watch Gilbert strang course too

When I was in school these were called "Simultaneous Equations". Not sure when the term "Systems of Equations" became the preferred term. I only heard for the first time about a year ago.

It was great seeing an extraneous solution that didn't solve the systems of those nonlinear equations by trying to turn a 3x2 matrix multiplied by a 2x1 matrix to form a 3x1 matrix of solutions. We saw extraneous x=1 that didn't work with x=6 and x=12 that both do work as solutions in what was taught. Sometimes looking at a problem creating polynomial solutions both extraneous and good solutions get intermixed to allow us to see potentials that fail but can inter result. 👍

Before looking at BPRP's solution: I was able to factor the information in the equations into a quadratic in terms of x, yielding two results. One case yielded a solution in the Rationals while the other yielded a solution in the Naturals, so depending on the set the variables are meant to be from, there are either one or two solutions. After watching BPRP's solution: the quadratic I obtained was the same, but the factor of (x - 1) in the cubic was a redundant solution because of the matrix multiplication. I eliminated that solution because I had used a linear transformation while I was manipulating the original equations to extract the quadratic in x. Always worth checking solutions with the original equations as well as any obtained through linear algebra manipulation to ensure they are true solutions rather than redundant (i.e contradictory) solutions.

Subtracting Eq 3 from Eq 1 and adding 1: xy-x-y+1=(x-1)(y-1)=22 Subtracting Eq 2 from Eq 1 and adding 1: xz-x-z+1=(x-1)(z-1)=33 Substituting X=x-1,etc., we get: XYZ=22Z=33Y Or Z=3Y/2 X+Y+Z=16. So 2X+5Y=32 Subtracting Eq 1 from Eq 2: zX-yX=(Z-Y)X=11. So XY=22 Now we get: (2X-5Y)^2 = (2X+5Y)^2 - 40XY = 1024 - 880 = 144 So 2X-5Y=+/- 12 4X = 44,20. We then get X, Y, Z, and finally get the two solutions: x=12,y=3,z=4 or x=6,y=27/5,z=38/5

This question is to be answered until 11:59 am, for those who have the last digit of their DNI 5: What is the letter you like the most among A, B, C, I, N , The m?

Set up the equation x(Eq3) + (Eq3), and simplify using (Eq1) and (Eq2), you get x^2 -18x +72 = 0, then solve for x. Next, multiply (Eq1) by x, simplify using (Eq2), solve for y to get y=(40x-51)/(x^2-1) and solve for y. Lastly use (Eq3) to solve for z

7:43 You can immediately see there's a X-1 factor in the determinant and factor it right away before expanding the 19-X. X²(19-X)+51+40-(19-X)-51X-40X = (X²-1)(19-X)+51(1-X)+40(1-X) = (X-1)(X+1)(19-X)-91(X-1) = (X-1)((X+1)(19-X)-91) = (X-1)(-X²+18X-72)

This is a beautiful problem.

3, 4, and 12 This is a simpleproblem for an Harvard MIT math tournament xy+ z =40 equation 1 xz + y =51 equation 2 x+ y+z = 19 equation 3 Let's add the first two equations : xy + z =40 and xz + y =51, Hence xy + xz + y + z =91 x(y+z) + 1(y+z) =91 factor out y +z Equation 5 y + z = 19- x (solving for y+ z , using equation 3) x(19-x) + 1 (19-x) = 91 (substituting 19-x into equation 5) (x+1)(19-x) =91 (x+1)(-x+19)=91 - x^2 + 19- x + 19x =91 0 = x^2 -18x +72 0= (x -6)(x-12) x =6 and x =12 Let try x=12 first using equation 1 and equation 2 12y + z=40 y + 12z=51 y = 3, and z=4 Hence x=12 , y=3 and z= 4 The next step is to solve when x =6 and using equation 1 and equation 2 again 6y + z =40 equation 9 y + 6z= 51 equation 10 36y + 6z = 240 mulltiply equation 9 by 6 y + 6z = 51 35y = 189 y= 189/35 y=5.4 z=266/35 z=7.6 x =6, y=5.4 , and z=7.6 answer as well This also satisfy the equation when x= 6, Hence x=12 Hence the answer is x=12, y=3 and z=4

we can solve by adding the equations xy+z=40 eq1 xz+y=51 eq2 x+y+z=19 eq3 on adding eq. 1,2,3 x+xy+xz+2y+2z=110 adding 2 on both the sides x(y+z+1) 2(y+z+1)=112 (x+2) (y+z+1)=112 -------equation 4 from equation 3 y+z=19-x substituting y+z=19-x in eq. 4. (x+2) (-x+20)=112 -x^2+18x-72=0 (x-6)(-x+12)=0--------equation 5 from equation 5 x=6, x=12 substituting x=6 in equations 1,2,3 we get 6y+z=40--eq6 6z+y=51---eq7 6+y+z=19 --eq8 from equation 8 y+z=13 from equation 6 z=40-6y substituting z=40-6y in eq. 8 -5y=-27 y=5.4 substituting y=5.4 in equation 8 z=7.6

i was thinking of row reduction while treating x as const

We can also solve it without using determinant. For example, let us arrange these equations as xz + y = 51 ...(1), xy + z = 40 ...(2) and x + y + z = 19 ...(3). Adding (1) and (2) and factorizing, we get (x + 1)(z + y ) = 91 ... (4). Again subtracting (2) from (1), we get , on factorizing, (z - y)(x - 1) = 11 ...(5). From (3), (z + y) = 19 - x ...(6). Putting (z + x ) value in (4) , we get (x + 1)(19 - x) = 91 ... (7) solving this , we get ( x - 6)(x -12) =0, so that we have x = 6 or x = 12. Case - I, when x = 6, in (3), we we get z + y = 13 ... (7) and again putting x =6 in (5), we get (z - y) = 11/5 ... (8). From (7) and (8) we get z = (1/2)(13+11/5) = 7.6 and y = (1/2)(13 - 11/5) = 5.4. Therefore, the first solution set {x, y, z) = {6, 5.4, 7.6}. Case - II when x =12: putting this in (3) z + y = 7 ...(9). Again putting x = 12 in (5) we get z -y = 1 ...(10). From (9) and (10), we get z = (1/2)( 7 + 1) = 4 and y = (1/2)( 7 - 1) = 3. Therefore, the 2nd solution set {x, y, z) = {12, 3, 4}.

11:55 that transition

I multiplied the 3rd by x, and then substituted xy and xz with the first 2 equations and then substituted y+z with the 3rd equation to get a quadratic equation with only x.

What is this sorcery dividing by x-1 can you explain that technique????

Ever heard of the Moore Penrose inverse

Is there a better way to check if the solutions are valid? Or is the only alternative to testing?

What questiined me is why did value 1 appeared? After a thought, beacuse it makes matrix singular not with row 3 being linear combination of rows 1 and 2, but columns 1,2 became collinear.

First eq take last eq: xy - x - y = 21 (IV) Rearrange third z = 19-x-y Sub this into second x(19-x-y) + y = 51 (V) Expand this and add to (IV) 19x - x^2 -xy +y +xy -x -y = 21+51 Simplify to x^2 -18x +72 =0 (x-6)(x-12)=0 x=6, or x=12 Then sub into (IV) to get y and then into third eq to get z. (6,5.4,7.6) and (12,3,4)

Nice. I didn't know that trick.

(40 + 51) / (x + 1) = 19 - x

Pretty interesting PROF 👍

just add equation first and second and factorise

Nice and very easy

It is a really good question

det(x 1, 1 x)=x^2-1 detψ(40 1,51 x ) =40x-51 detz(χ 40,1 51)=51χ-40 ψ=detψ /det z=detz/det we substitute into third equation x^3-19x^2 +90x-72=0 x1 x=6 ψ=5.4 z=7.6 x=12 ψ=3 z=4 x>=1 x

I guess I over-complicated this: xz + y - (x - y - z) = 32 => xz - x - z +1 = 33 => (x-1)(z-1) = 33 (1) xz + y - xy - z = 11 => (x-1)(z-y) = 11 (2) 1/2 => z - 1 = 3(z - y) => z = (3y - 1)/2 => x = (39 - 5y)/2 xy + z = 40 => 5y^2 - 42y + 81 = 0; y = 3 or y = 27/5 (x,y,z) = (6, 27/5, 38/5) or (12, 3, 4)

Just subtract eq.3 from eq.2 = eq.4 Subtract eq.4 from eq.1=eq.5 Substitute the value of x from eq.3 in eq.5 Factorise eq.5 and you get value of x Substitute value of x and y in eq.1 to get y and z

Could also use multivariable Newton method

So, we're basically treating the X as a parameter and using the standard method for linear system? Well, pretty genious move; good job americans

Truely useful method that gave me new insights about using the traditional Gaussian Elimination However, If I were to be in that tournament, i would use the traditional Substitution method, because it seems specifically easier and faster in this question

xy+(19-x-y)=40 x(19-x-y)+y=51 adding both eq to get 18x-x^2+19=91 so x^2-18x+72=0 then x=6, x=12 solve for y and z and get 2 solutions 12,3,4 and 6,27/5,38/5

You could have just added equations 1 and 2 and factorise as (x+1)(y+z) = 91. Then substitute from equation 3 y+z = 19-x…..

Esta pregunta es para que me respondan hasta las 11:59 am, para los que tienen el último dígito de su DNI 5 : ¿Cuál es la letra que les gusta más entre la A, la B, la C, la I, la N, la M?

Basically, we assume that this system is solvable, and from that assumption we solving simpler system, avoiding undefined behaviour, like setting variable to zero.

I haven't seen Ruffini in ages!

Did it in my head, 12,4,3

12,3,4

Could you explain why cos(sin(1/e)) ≈ 1? (In degrees)

I have a question, Let A B C, and k are Real numbers. If 2k^2k = ((A+B+C)^(A+B+C))/2, find k in terms of A, B, and C.

I have a type of equation I'm curious about x!=a how can we easily solve this for any given value of a I couldn't find anyone even mention factorialic equations like this I'm sure there is a solution since we can just guess it by insertion or solve it graphically but that's lame so is there a better way?

Dude I have similar question for you to solve x+y+z=4, xy+yz+zx= -7, xyz= -10 find x, y & z. tell me in another way or easy way for this question i try substitution but it was became very long and lengthy.

12,4,3

I assumed that x had to control the rank of the system. I determined that the rank had to be 2, so I did Converted the equation to augmented matrix system (4 x 3) and determined the x value to ensure that one of the rows go to 0’s for rank 2. I got I think 2 values for x (I have to check I did this a long time ago) that trip this equation to a rank 2. Then solved for y a and x using cramers rule. Anyways moral of the story: I don’t think you have to get the characteristic polynomial to trip this into a diminished rank anyways.

Solved this in the real test! :)

The part you kinda zipped through was that cube root part? You said assume 1 is a sol. and then proceeded with that approach I’m not familiar with

I just think of an easy way which high school can solve this. Since x +y +z = 19, z = 19 -x -y ; xy +z = 40, so xy + 19 -x -y = 40, then xy -x -y =21 … (1) xz + y = 51, so x (19 -x -y) +y = 51, then 19x -x^2 -xy +y = 51 …(2) Put (2) into (1), -x^2 +19x -xy +x +y -x =51, so -x^2 +18x -21 =51, x^2 -18x +72 = 0. We can get x = 6 or 12

Mit never disappoints 🗿

This man is a genius

If we take our solutions to be in projective space, the x=1 case provides one more solution at infinity: (1, 0, 0, 0).

Surprisingly Seeing the third eqn first time i randomly put values (x,y,z) as (12,3,4) and ❤ question is over But answer is not the thing that mathematics wants😊 Now trying with the original method

add all eqns , we get xy + z + xz + y + x + y + z = 110 xy + xz + x + 2y + 2z = 112 - 2 x(y + z + 1) + 2y + 2z + 2 = 112 x(y + z + 1) + 2(y + z + 1) = 112 (y + z + 1)(x + 2) = 112 (y + z + 1)(x + 2) = 8 * 14 x + 2 = 14, y + z + 1 = 8 x = 12..........(a), y + z = 7.........(b) given xy + z = 40 from eq(a), 12y + z = 40 11y + (y + z) = 40 from eq(b), 11y = 40 - 7 11y = 33 y= 3.........(c) also given, x + y + z = 19 from eq(a) and (c), 12 + 3 + z = 19 z = 4 (x,y,z) = (12,3,4) integer soln

Add first two equations and put y+z = 19-x

I wonder if solving geometrically in 3d is any better

Me using Newton’s method to approximate the solutions 😢

Very nice method, but i have a confusion like why this method work ? You get the value of x such that it make 1 equation redundant and i got that , but this doesn’t explain why such X value will be a value that satisfies the original equation with the other variables?

The method above makes absolutely no sense to me. I solved the bottom equation for y: y=19-x-z Then I substituted it into the first two equations which yields the following two equations in terms of x and z: -x^2+19x-xz+z=40 -x+xz-z=32 When you add these two equations together, the z and xz terms cancel out and you're left with the quadratic: x^2-18x+72=0, so x=6 or 12. From there, it is easy to solve for y and z. (x,y,z) = (6, 27/5, 38/5) or (12, 3, 4).

No can you do non-linear equations? Like x^2 + x*y*z + y^2*z = 50 and higher powers of x and y and z. Or even weirder ones like "sin(x) + y = 30" and "x + sin(y) = 40"?