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If $\mathrm{x}, \mathrm{y}, \mathrm{z}$ are different from zero and
$\Delta=\left|\begin{array}{ccc}a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c\end{array}\right|=0$ then the value of
the expression $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$ is
Options:
$\Delta=\left|\begin{array}{ccc}a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c\end{array}\right|=0$ then the value of
the expression $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}$ is
Solution:
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The correct answer is:
2
$\left[\begin{array}{lll}a & b-y & c-z \\ a-x & b & c-z \\ a-x & b-y & c\end{array}\right]=0$
$\Rightarrow a(b c-b c+b z+c y-y z)-(b-y)$
$(a c-c x-a c+a z+c x-z x)+(c-z)$
$(a b-a y-b x+x y-a b+b x)=0$
$\Rightarrow a b z+a c y-a y z-a b z+b z x+a y z-$
$\Rightarrow a y z+b z x+c x y-2 x y z=0$
$\Rightarrow \frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2=0$ (Dividing by $\left.x y z\right)$
$\Rightarrow \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$
$\Rightarrow a(b c-b c+b z+c y-y z)-(b-y)$
$(a c-c x-a c+a z+c x-z x)+(c-z)$
$(a b-a y-b x+x y-a b+b x)=0$
$\Rightarrow a b z+a c y-a y z-a b z+b z x+a y z-$
$\Rightarrow a y z+b z x+c x y-2 x y z=0$
$\Rightarrow \frac{a}{x}+\frac{b}{y}+\frac{c}{z}-2=0$ (Dividing by $\left.x y z\right)$
$\Rightarrow \frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$
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