Question

The number of real values of $a$ for which the system of equations
$x+ay-z=0,2x-y+az=0,ax+y+2z=0$
has a non-trivial solution, is

• A. 3
• B. 1
• C. 0
• D. infinite

Answer 1

Answer: (A)

Since the given system of equations has a non-trivial solution,
$\Delta =\left|\begin{array}{ccc}1& a& -1\\ 2& -1& a\\ a& 1& 2\end{array}\right|=0$
Using $C_1\to C_1+C_3,C_2\to C_2+a{C}_3$, we get
$\Delta =\left|\begin{array}{ccc}0& 0& -1\\ 2+a& -1+a^2& a\\ 2+a& 1+2a& 2<br/>\end{array}\right|=0$
$\Rightarrow (-1)\left|\begin{array}{cc}2+a& -1+a^2\\ 2+a& 1+2a\end{array}\right|=0$
$\Rightarrow (2+a)\left(1+2a+1-a^2\right)=0$
$\Rightarrow a=-2,1\pm \sqrt{3}$ .
Thus, there are three real values of $a$ for which the system of equations has a non-trivial solution.