The determinant of the matrix:
⎡1/a bc b+c⎤
⎢1/b ac a+c⎥
⎣1/c ab a+b⎦
is zero, as we'll see below.
Take the third column. Its role in the determinant is given by solving the 2×2 determinant:
⎢1/b ac⎥
⎢1/c ab⎥
The diagonal products result in a-a=0, so the third column contributes zero to the evaluation of the 3×3 determinant.
What about the other two columns' contributions? Consider the first column which uses the 2×2 determinant:
⎢ac a+c⎥
⎢ab a+b⎥
First, the factor a in the first column of this determinant can be extracted so that we have effectively:
⎢c a+c⎥
⎢b a+b⎥
multiplied by scalar factor a.
The product bc is eliminated in the diagonal product, leaving us with just ac-ab which can also be factorised: a(c-b). The result is a2(c-b), which is not zero. When this is multiplied by element 1/a we end up with just a(c-b).
Now for the contribution of the second column. The actual arithmetic would be -bc((a+b)/b-(a+c)/c). Conveniently the denominators of the fractions both happen to be in the multiplier bc. So we end up with -c(a+b)+b(a+c) in which, again, the product bc is eliminated. This leaves us with -ca+ab=a(-c+b), common factor a again. Magically, this cancels the residue a(c-b) we found earlier so the determinant of the 3×3 matrix is zero.