For each real number α, let $B_α$ be the bilinear form $B_α((x_1, y_1, z_1),(x_2, y_2, z_2)) =x_1x_2 + 2y_1y_2 + 11z_1z_2 + (α + 1)(x_1y_2 + y_1x_2) + 3(x_1z_2 + z_1x_2) + (α + 2)(y_1z_2 + z_1y_2).$
Find the set {$α ∈ R : B_α$ is positive definite}
My attempts : i was reading this articles..but couldnot get any hints to tackle this question.......
http://www.imsc.res.in/~knr/ece15/ps.pdf
my thinking : i take $x_1= x_2 , y_1= y_2, z_1 =z_3$...after that i make
$B_α((x_1, y_1, z_1),(x_1, y_1, z_1)=x_1x_1 + 2y_1y_1 + 11z_1z_1 + (α + 1)(x_1y_1 + y_1x_1) + 3(x_1z_1 + z_1x_1) + (α + 2)(y_1z_1 + z_1y_1).$
$B_α((x_1, y_1, z_1),(x_1, y_1, z_1)=x_1^2 + 2y_1^2 + 11z_1^2+ (α + 1)(2x_1y_1 ) + 3(2x_1z_1 ) + (α + 2)(2y_1z_1 ).$
Now converting them into matrix
$$B_α =\begin{bmatrix} 1& α+1 &3 \\ α+ 1 &2 & α+2\\ 3& α +2 & 11\end{bmatrix} >0$$
For positive definite $Det B > 0$
= $-α^2 - 2α + 7 -8(α^2 + 13α + 5) +3α^2 + 3α -12 > 0 $
=$-6α^2 -12α > 0$
= $α > - \frac {12}{6}$
=$α > - 2$
Is my answer is correct ? or not correct ???
Pliz verified my answer