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Math Help - Homogeneous equations

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by nahduma View Post
    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!
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  3. #3
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    But what is the answer?

    I understand that, but it still doesn't answer my question.
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  4. #4
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    <br />
A=\begin{bmatrix}<br />
a&1&1 \\<br />
 1&b&1 \\<br />
 1&1&c \\<br />
 \end{bmatrix},<br />
\det(A)=(a-1)[c(b-1)-(1-c)]+(1-b)(1-c)<br />

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