System of linear equations

[fredreload]

So I got this name from Wikipedia. Basically I got a few linear equations of n equations n unknowns so that it is always solvable. For example.

 

val=[3,1,2,3,4,5,3,1,8,7,9,5,0,2,5,3]

 

I got by splitting in half x=[3,1,2,4,6,5,3,1], y=[8,7,9,5,0,2,5,3],z=[?,?,?,?,?,?,?,?],w[?,?,?,?,?,?,?,?]

 

So for each x and y value I plug it into the equation to get z and w

2x+y=14

2x-y=-2

 

x=3, y=8

z=14,w=-2

 

 

Now if I put this in Matrix form it would be

|2 1|14|

|2 -1|-2|

 

Which I can reduce this matrix with the row reduction formula to solve for x and y using z=14 and w=-2 with the constants

|1 0|3|

|0 1|8|

 

My point is you can control the constants however you see fit, meaning you can change the equation to

3x+y=17

3x-y=1

You would still get x=3 and y=8 except now z=17 and w=1 so the matrix become.

|3 1|17|

|3 -1| 1|

The equation can also be in polynomials, derivatives, anti-derivatives, as long as I can back solve x and y from w and z. And linear equation is the easiest solvable one.

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But let me show you a mistake I made when I tried to reduce the formula.

For instance, I set

x-y=-5

y-x=5

These two equations looks opposite of each other, and by doing this I've successfully reduced the output z and w where they are just the opposite of each other z=-w, but this equation doesn't work because they are actually the same equation, so it's 1 equation 2 unknowns and you cannot solve for x and y with this

x-y=5

y-x=5, (-1)(x-y)=5, x-y=-5

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So I want a way to reduce the z and w outputs maybe so that they can reference to each other, but still able to solve for x and y.

I can also split val into 3 parts or 4 parts and introduce another variable into the equation for instance

val=[3,1,2,3,4,5,3,1,8,7,9,5,0,2,5,3]

v=[3,1,2,3],w=[4,5,3,1],x=[8,7,9,5],y=[0,2,5,3] and 4 results [?,?,?,?],[?,?,?,?],[?,?,?,?],[?,?,?,?]

So now you have 4 equations 4 unknowns

v+w+x+y=15

v+w+x-y=15

v+w-x+y=-1

v-w+x+y=7

| 1 1 1 1|15|

| 1 1 1 -1|15|

| 1 1 -1 1|-1|

| 1 -1 1 1| 7|

And at this point if you still knows what I'm talking about, tell me how to set up the equations constants, so I can reduce the results = =

Edited September 5, 2016 by fredreload

[fredreload]

This one is attributed to League of Legends, Google, and Shokugeki no Soma and free for use

Edited September 5, 2016 by fredreload

[Country Boy]

I am simply not clear what you are doing! The system of equation [math]a_{n1}x_1+ a_{n2}x_2+ \cdot\cdot\cdot+ a_{nn}x_n= z_n[/math] has a unique solution as long as the determinant of coefficients is non-zero, irrespective of what numbers you have on the right side. In your first example, that determinant is [math]\left|\begin{array}{cc}2 & 1 \\ 2 & -1\end{array}\right|= 2(-1)- 1(2)= -4[/math], which is non-zero. In the second, it is [math]\left|\begin{array}{cc} 3 & 1 \\ 3 & -1\end{array}\right|= 3(-1)- 1(3)= -6[/math], non-zero. For the third example, x- y= 5 and y- x= -5, which is the same as 1(x)- 1(y)= 5, -1(x)+ 1(y)= -5, the determinant is [math]\left|\begin{array}{cc} 1 & -1 \\ -1 & 1\end{array}\right|= 1(-1)- (-1)(1)= 0[/math] so there is no unique solution. But it is easy to see that if you multiply both sides of x- y= 5 by -1 you get -x+ y= y- x= -5. That is you really have only one equation. There are an infinite number of solutions- pick any number you like for x and take y to be x- 5. You can, by taking w to be -z, get an infinite number of solutions but if [math]w\ne -z[/math], there is no solution.

[fredreload]

Hmm, no unique solution noted