Given $x=c y+b z, y=a z+c x, z=b x+a y$ where $x, y, z$ are not all zero, then $a^{2}+b^{2}+c^{2}+2

Question

Given $x=c y+b z, y=a z+c x, z=b x+a y$ where $x, y, z$ are not all zero, then $a^{2}+b^{2}+c^{2}+2 a b c=$

Professor PC Thomas Classes · Answer

According to question, $(-1) x+c y+b z=0$ $c x+(-1) y+a z=0, b x+a y+(-1) z=0$\ Above is a system of three linear homogeneous equations and since $x, y, z$ are not all zero, it has a nontrivial solution. So we must have coefficient determinant must be zero.\ $	herefore\left|\begin{array}{ccc}-1 & c & b \ c & -1 & a \ b & a & -1\end{array}ight|=0$\ Expanding $(-1)\left(1-a^{2}ight)-c(-c-a b)$ $+b(c a+b)=0$\ $\Rightarrow \quad a^{2}+b^{2}+c^{2}+2 a b c-1=0$\ $\Rightarrow a^{2}+b^{2}+c^{2}+2 a b c=1$