JEE MAIN - Mathematics (2008 - No. 9)
Let $$a, b, c$$ be any real numbers. Suppose that there are real numbers $$x, y, z$$ not all zero such that $$x=cy+bz,$$ $$y=az+cx,$$ and $$z=bx+ay.$$ Then $${a^2} + {b^2} + {c^2} + 2abc$$ is equal to :
$$2$$
$$-1$$
$$0$$
$$1$$
Explanation
The given equations are
$$\matrix{ { - x + cy + bz = 0} \cr {cx - y + az = 0} \cr {bx + ay - z = 0} \cr } $$
As $$x,y,z$$ are not all zero
$$\therefore$$ The above system should not have unique (zero) solution
$$ \Rightarrow \Delta = 0 \Rightarrow \left| {\matrix{ { - 1} & c & b \cr c & { - 1} & a \cr b & a & { - 1} \cr } } \right| = 0$$
$$ \Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$$
$$ \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$$
$$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$$
$$\matrix{ { - x + cy + bz = 0} \cr {cx - y + az = 0} \cr {bx + ay - z = 0} \cr } $$
As $$x,y,z$$ are not all zero
$$\therefore$$ The above system should not have unique (zero) solution
$$ \Rightarrow \Delta = 0 \Rightarrow \left| {\matrix{ { - 1} & c & b \cr c & { - 1} & a \cr b & a & { - 1} \cr } } \right| = 0$$
$$ \Rightarrow - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$$
$$ \Rightarrow - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$$
$$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$$
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