Question

The number of real values of $a$ for which the system of equations
$x+a y-z=0,2 x-y+a z=0, a x+y+2 z=0$
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,
$\Delta=\begin{vmatrix}1 & a & -1 \\2 & -1 & a \\a & 1 & 2\end{vmatrix}=0$
Using $C_1 \rightarrow C_1+C_3, C_2 \rightarrow C_2+a C_3$, we get
$ \Delta=\begin{vmatrix}0 & 0 & -1 \\2+a & -1+a^2 & a \\2+a & 1+2 a & 2
\end{vmatrix}=0 $
$\Rightarrow (-1)\begin{vmatrix}2+a & -1+a^2 \\2+a & 1+2 a\end{vmatrix}=0 $
$\Rightarrow (2+a)\left(1+2 a+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.