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

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[nzbil]

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

[DistortedV]

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

[harshitgautam]

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

[Steve_Stowers]

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.

[landsgevaer]

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.

[kontiki]

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)

[lostwizard]

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. :)

[zycnthropy]

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

[victorpaesplinio]

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

[Chris_]

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

[matheusjahnke]

Time for cheap tricks;
Computing L1+L2-L3- x L3
on the left hand side... we have

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

[khattab]

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)

[felipefred]

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

[joshuanugentfitnessjourney]

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

[eqyuio]

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

[sanjogar]

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.

[iabervon]

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.

[TalkLoudSayNothing]

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.

[devondevon]

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

[jimschneider]

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?