Discussion Forum

If the equations x + ay - z = 0, 2x - y + az = 0, ax + y + 2z = 0 have non-trivial solutions, then a =

Answer:

Explanation:

$\begin{vmatrix}1 & a & -1 \\2 & -1 & a\\ a & 1& 2\end{vmatrix} = 0$

Applying R₂→ R₂ - 2R₁ and R₃ → R₃ - aR₁, we get

$\begin{vmatrix}1 & a & -1 \\0 & -1-2a & a+2\\ 0 & 1-a^{2}& 2+a\end{vmatrix} = 0$

Expanding along C₁, we get

$(a + 2) (a^{2} - 2a - 2) = 0\Rightarrow  a= - 2, 1\pm \sqrt{3}$