For a real number α, if the system [ cases 1 & α & α ^ 2 α & 1 & α...

Determinant value of 3$\times$3 matrix should be zero for infinite solution

$\Delta = 0 ⇒ 1\left(1 - {\alpha}^{2}\right) - \alpha\left(\alpha - {\alpha}^{3}\right) + {\alpha}^{2}\left({\alpha}^{2} - {\alpha}^{2}\right) = 0$

\left(1 - {\alpha}^{2}\right) - {\alpha}^{2} + {\alpha}^{4} = 0$</p><p>${\left({\alpha}^{2} - 1\right)}^{2} = 0 ⇒ \alpha = \pm 1$</p><p>But at$\alpha = 1$, three equations become parallel, so no solution so rejected</p><p>at$\alpha = - 1$, All three equation become</p><p>$x - y + z = 1$(coincident planes), so infinite solution.</p><p>$∴ 1 + \alpha + {\alpha}^{2} = 1