I need help with this exercise:
Show that the matrix B =\begin{pmatrix} 3 & 2 & 1 \\ 2 & 4 & 1 \\ 1 & 1 & 1 \end{pmatrix}
has exactly one eigenvalue in the open interval (1,2) by studying the signatures of the quadratic forms with matrices $B-I$ and $B-2I$.
The only thing I have managed to do is compute the signatures of $B-I$ and $B-2I$. They're (2,1) and (1,2) respectively. I don't know how the open interval comes into play.
Hint: For a scalar $\alpha$, the eigenvalues of $B - \alpha I$ are of the form $\lambda_i - \alpha$ for each eigenvalue $\lambda_1,\lambda_2,\lambda_3$ of $B$. On the other hand, the signatures tell you that $B - (1)I$ has $2$ positive eigenvalues while $B - 2I$ has $1$ positive eigenvalue.