Question

The real value of 'a' for which the equations ax+y+z=0, -x+ay+z=0, -x-y+az =0 have nonzero solution is

• A. 1
• B. 0
• C. -1
• D. all the above

Answer 1

The correct option is B 0

Given, $ax+y+z=0,-x+ay+z=0,-x-y+az=0$

$\therefore \Delta =\left|\begin{array}{ccc}a& 1& 1\\ -1& a& 1\\ -1& -1& a\end{array}\right|$
$=a^3+a+a-1+1+a$
$=a^3+3a$
$\therefore a^3+3a=0$
$\therefore a(a^2+3)=0$
$\therefore a^2+3\ne 0$

$\therefore a=0$