Question

Given, $x=cy+bz,y=az+cx,z=bx+ay$, where x, y, z are not all zero, and a, b, c are real numbers. The value of $a^2+b^2+c^2+2abc$ is

• A. \N
• B. 4
• C. 2
• D. 1

Answer 1

The correct option is D 1

Given systems of equations can be rewritten as
$-x+cy+by=0,cx-y+az=0andbx+ay-z=0$
Above system of equations are homogeneous equation. Since, x, y and z are not all zero, so it has non - trivial solution.
Therefore, the coefficient of determinant must be zero.
$\therefore \left|\begin{array}{ccc}-1& c& b\\ c& -1& a\\ b& a& -1\end{array}\right|=0$
$\Rightarrow -1(1-a^2)-c(-c-ab)+b(ca+b)=0$
$\Rightarrow a^2+b^2+c^2+2abc-1=0$
$\Rightarrow a^2+b^2+c^2+2abc=1$