GeckoFood's Journal: [fun] Mathematical challenge

Given that you have a, b, c, and d:

W = a + b
X = b + c
Y = c + d
Z = d + a

If you have W, X, Y and Z already, how do you solve for a, b, c, and d? Not quite as simple as it appears...

Barbie says math is hard

[bofkentucky]

It works if all the constants are 0, other than that I can't wrap my head around it today.

Linear Algebra

[RailGunner]

Convert it into a Matrix and reduce.

How to solve these. [wikipedia.org]

/not doing the work here
// Tigers down 10-0 against the f-ing Rangers has me in a bad mood.

Re:

[GeckoFood]

That makes for an interesting reduction exercise... Every row always has 2 values in it:

1 1 0 0 | W
0 1 1 0 | X
0 0 1 1 | Y
1 0 0 1 | Z

So, while you can get rid of one value in one column, you introduce a new value in a different column.

Re:

[Morosoph]

1 1 0 0 | W
0 1 1 0 | X
0 0 1 1 | Y
1 0 0 1 | Z

The matrix is singular and cannot be solved.

Det (M) = Det
{1 1 0
0 1 1
0 0 1}
- Det
{1 0 0
1 1 0
0 1 1} = 1 - 1