Compute eigenvalues of Hessian = $\begin{bmatrix}a&1\\1&2\end{bmatrix}$ such that function is convex/eigenvalues $\geq 0$ The Hessian matrix is given to be $\begin{bmatrix}a&1\\1&2 \end{bmatrix}$ where $a$ is a real number.
EDIT: So to find the eigenvalues I find the determinant which is ad-bc so 2a-1.
So in order for the function to be convex $a\geq\frac{1}{2}$

Your answer is correct.
However: you the determinant of the matrix is not $(\lambda - a)(\lambda -2)-1$. 
The determinant of the matrix ${\bf M}$ is $2a-1$.
You found, quite rightly, the determinant $\det(\lambda\bf{I} - {\bf M})$.
