Suppose a,b,c ∈Rand abc,α≠$$0$$ If the system of $$equationsa+$$α$$x+$$α$$y+$$α$$z=0-----(1)$$α$$x+b+$$α$$y+$$α$$z=0-----(2)$$α$$x+$$α$$y+$$α$$+cz=0------(3)$$ has a $$non-trivial$$ solution, then α$$1a+1b+1cis=$$

Question

Suppose a,b,c ∈Rand abc,α≠$$0$$ If the system of $$equationsa+$$α$$x+$$α$$y+$$α$$z=0-----(1)$$α$$x+b+$$α$$y+$$α$$z=0-----(2)$$α$$x+$$α$$y+$$α$$+cz=0------(3)$$ has a $$non-trivial$$ solution, then  α$$1a+1b+1cis=$$

Infinity Learn · Accepted Answer

From eqns $$1$$ &$$2$$ we haveax−$$by=0$$⇒$$ax=byFrom$$ eqns $$2$$ &$$3$$ we haveax−$$cz=0$$⇒$$ax=czax=by=cz$$⇒$$x/1/a=y/1/b=z/1/cSubstituting$$ x,y,z values in equation (1), we get $$a+$$α$$a+$$α$$b+$$α$$c=0=1$$ ⇒ α$$1a+1b+1c=$$−$$1$$

Suppose a,b,c $\in R$ and $a b c, α \neq 0$ If the system of equations
$(a + α) x + α y + α z = 0 - - - - - (1)$
$α x + (b + α) y + α z = 0 - - - - - (2)$
$α x + α y + (α + c) z = 0 - - - - - - (3)$ has a non-trivial solution, then $α (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) i s =$

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Suppose a,b,c ∈Rand abc,α≠0 If the system of equationsa+αx+αy+αz=0-----(1)αx+b+αy+αz=0-----(2)αx+αy+α+cz=0------(3) has a non-trivial solution, then α1a+1b+1cis=

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Ready to Test Your Skills?

free study material

Suppose a,b,c $\in R$ and $a b c, α \neq 0$ If the system of equations
$(a + α) x + α y + α z = 0 - - - - - (1)$
$α x + (b + α) y + α z = 0 - - - - - (2)$
$α x + α y + (α + c) z = 0 - - - - - - (3)$ has a non-trivial solution, then $α (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) i s =$