Question

Let a,b and c be any real numbers. Suppose that there are real numbers $x,y,z$ not all zero such that $x=cy$, $y=az+cx$ and $z=bx+ay$, then $a^2+b^2+c^2+2abc$ is equal to:

• A. $2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is D $1$

System of equations

$x-cy-bz=0$

$cx-y+az=0$

$bx+ay-z=0$

has non trivial solution if the determinant of coefficient matrix is zero

$\Rightarrow \left|\begin{array}{ccc}1& -c& -b\\ c& -1& a\\ b& a& -1\end{array}\right|=0$

$\Rightarrow 1(1-a^2)+c(-c-ab)-b(ca+b)=0$

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

$\Rightarrow -a^2-b^2-c^2-2abc=-1$

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