Homogeneous Equations

Algebra Level 4

( a + α ) x + α y + α z = 0 α x + ( α + b ) y + α z = 0 α x + α y + ( α + c ) z = 0 (a+\alpha )x+\alpha y+\alpha z=0\\ \\ \alpha x+(\alpha +b)y+\alpha z=0\\ \\ \alpha x+\alpha y+(\alpha +c)z=0

Given that a,b,c are non-zero real numbers satisfying the system of equations above.

The given system of equations have a non-trivial solution if α − 1 = k ( a − 1 + b − 1 + c − 1 ) { \alpha }^{ -1 }=k({ a }^{ -1 }+{ b }^{ -1 }+{ c }^{ -1 }) then what is the the value of 'k'?


The answer is -1.

Sahil Bansal by Sahil Bansal , 21, India

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1 solution

Harsh Khatri
Apr 8, 2016

After rearranging the given equations, we get:

α ( x + y + z ) = − a x = − b y = − c z \alpha (x+y+z) = -ax = -by = -cz

Using α ( x + y + z ) = − c z \alpha(x+y+z) = -cz and x = c a z ; y = c b z x=\frac{c}{a} z; y=\frac{c}{b}z , we get:

α ( c a + c b + 1 ) z = − c z \alpha( \frac{c}{a} + \frac{c}{b} + 1)z = -cz

Dividing throughout by c z cz :

⇒ α ( 1 a + 1 b + 1 c ) = − 1 \displaystyle \Rightarrow \alpha( \frac{1}{a} + \frac{1}{b} + \frac{1}{c}) = -1

⇒ α − 1 = ( − 1 ) × ( a − 1 + b − 1 + c − 1 ) \displaystyle \Rightarrow \alpha^{-1} = (-1) \times (a^{-1} + b^{-1} + c^{-1})

Hence, k = − 1 k = \boxed{-1}

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