Bounds on the determinant of a $3 \times 3$ matrix I have a matrix $$B = A + \epsilon I$$
where $A$ is a $3 \times 3$ matrix, $I$ is an identity matrix and $\epsilon$ is a positive scalar. 
Is there a lower bound on the determinant of $B$?
If A is a symmetric positive semi-definite matrix (say a co-variance matrix), is there a lower bound, when $\epsilon$ is a positive?
 A: The way your question is posed, there is no lower bound. Simply choose $A=(c-\epsilon)I$ with arbitrary $c$ and your matrix will have determinant $c^3$, which runs through $\mathbb{R}$, i.e. the determinant can take any value.
Or maybe, you want a lower bound in terms of the determinant of $A$? In that case, you can always just transform A into Jordan form (I won't be affected) and just read off the determinant. This should also (given some knowledge about $A$) give you lower bounds. 
For instance, it's clear that $\operatorname{det} B=0$ is always possible, regardless of the determinant of $A$ (you only need one entry $-\epsilon$ on the diagonal of $A$), so a lower bound must be $<0$. If the determinant is not zero, it will be a polynomial in $\epsilon$, which should give you some real bounds given some assumptions on $A$.
A: 
If A is a symmetric positive semi-definite matrix (say a co-variance matrix), is there a lower bound, when $\epsilon$ is a positive?

Given that the determinant is a product of all the eigenvalues of a matrix, that a positive semidefinite matrix has only nonnegative eigenvalues $\lambda_k(A) \ge 0$, that $\lambda_k(A + \epsilon I) = \lambda_k(A) + \epsilon$, and that $\epsilon \ge 0$, the lower bound is $\det A$ (for $\epsilon = 0$).
