Question

Find the Solution : $x-cy-bz=0$

$cx-y+az=0$

$bx+ay-z=0$

• A. $a^3+b^3+c^3+3abc=1$
• B. $a^2+b^2+c^2+2abc=1$
• C. $a^4+b^4+c^4+4abc=1$
• D. $a^5+b^5+c^5+5abc=1$

Answer 1

The correct option is B $a^2+b^2+c^2+2abc=1$

The given equations are

$x-cy-bz=0$
$cx-y+az=0$
$bx+ay-z=0$

Since x, y, z are not all zero, the system will have non-trivial solution if

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

or $1(1-a^2)+c(-c-ab)-b(ac+b)=0$

or $1-a^2-c^2-abc-abc-b^2=0$

or $a^2+b^2+c^2+2abc=1$