1. ## Homogeneous equations

If the system of 3 equations
ax+b+z=0
x+by+z=0
x+y+cz=0

has a non-trivial solution, then 1/(1-a) + 1/(1-b) + 1/(1-c) = ?

The choices are 1, 2, 0 or -1

It seems obvious to me that if a=b=c=1, the system reduces to one equation and has a non-trivial solution. But infinity isn't among the options.
Oh, and there isnt any clause stating that a, b and c are distinct.

2. Originally Posted by nahduma
If the system of 3 equations
ax+b+z=0
x+by+z=0
x+y+cz=0

has a non-trivial solution, then 1/(1-a) + 1/(1-b) + 1/(1-c) = ?

The choices are 1, 2, 0 or -1

It seems obvious to me that if a=b=c=1, the system reduces to one equation and has a non-trivial solution. But infinity isn't among the options.
Oh, and there isnt any clause stating that a, b and c are distinct.
You have 3 equations in 3 unknowns. Geometrically, this means that the 3 planes coincide and that those values of x,y,z that satisfy the equation ax+b+z=0, automatically satisfy all three equations. Most of the times, a system with the same number of equations and unknowns has a single unique solution. By the way, we say that a linear system is homogeneous if its constant term is zero!

3. ## But what is the answer?

I understand that, but it still doesn't answer my question.

4. $\displaystyle A=\begin{bmatrix} a&1&1 \\ 1&b&1 \\ 1&1&c \\ \end{bmatrix}, \det(A)=(a-1)[c(b-1)-(1-c)]+(1-b)(1-c)$

If rank(A)<3 then
$\displaystyle \det(A)=0 \\ \rightarrow (a-1)[c(b-1)-(1-c)]+(1-b)(1-c)=0 \\$
$\displaystyle \rightarrow (1-b)(1-c)=(1-a)[c(b-1)-(1-c)] \rightarrow \frac{1}{1-a}=\frac{c(b-1)-(1-c)}{(1-b)(1-c)}$
$\displaystyle \rightarrow \frac{1}{1-a}=\frac{c}{1-c}-\frac{1}{1-b} \\ \rightarrow \frac{1}{1-a}=1-\frac{1}{1-c}-\frac{1}{1-b} \\ \rightarrow \frac{1}{1-a}+\frac{1}{1-c}+\frac{1}{1-b}=1$