QJEEM08Mab225: JEE Main 2008 Mathematics - Class 12 JEEM08Mab225 Solution: Let be any real numbers. Suppose that there are real numbers…

Question

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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 :

Accepted Answer

The given equations are

\begin{matrix} - x + c y + b z = 0 \\ c x - y + a z = 0 \\ b x + a y - z = 0 \end{matrix}

As

x,y,z

are not all zero

\therefore

The above system should not have unique (zero) solution

\Rightarrow Δ = 0 \Rightarrow | \begin{matrix} - 1 & c & b \\ c & - 1 & a \\ b & a & - 1 \end{matrix} | = 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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