Let a, b, c be any real numbers.

Question

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{pmatrix} { - x + cy + bz = 0}  \
   {cx - y + az = 0}  \
   {bx + ay - z = 0}  \

\end{pmatrix} $$

As $x,y,z$ are not all zero

$$\therefore$$  The above system should not have unique (zero) solution

$$ \Rightarrow \Delta  = 0 \Rightarrow \left| {\begin{pmatrix} { - 1} & c & b  \
   c & { - 1} & a  \
   b & a & { - 1}  \

\end{pmatrix} } \right| = 0$$

$$ \Rightarrow  - 1\left( {1 - {a^2}} \right) - c\left( { - c - ab} \right) + b\left( {ac + b} \right) = 0$$

$$ \Rightarrow  - 1 + {a^2} + {b^2} + {c^2} + 2abc = 0$$

$$ \Rightarrow {a^2} + {b^2} + {c^2} + 2abc = 1$$

Solutions · 1