Invertibility of perturbed matrices

Question

Suppose I have two real symmetric matrices $A$ and $B$, where $B$ is invertible. $A$ may or may not be invertible. Both $A$ and $B$ have positive and negative eigenvalues (i.e., no positive semi-definiteness).

I'm interested in the invertibility of $A+\delta B$, where $\delta$ is a very small perturbation parameter. Intuitively, as $\delta$ gets arbitrarily small the fixed matrix $A$ can no longer "mess up" the eigenvalues of $A+\delta B$ and the perturbed matrix should be invertible. This leads me to the following question:

Question 1: Does there exist a small $\epsilon>0$ (which may depend on $A$ and $B$) such that, for all $\delta$ with $0<|\delta|<\epsilon$, the perturbed matrix $(A+\delta B)$ is invertible? Furthermore, is it true that $\|(A+\delta B)^{-1}\|_{op} \leq C(A,B)\cdot \delta^{-1}$ for small $\delta$, where $C(A,B)$ is some constant that only depends on $A$ and $B$?

A more general question involves multiple matrices:

Qeustion 2: Let $A,B_1,\cdots,B_r$ be fixed real symmetric matrices and suppose $B_r$ is invertible. Does there exist a small $\epsilon>0$ such that, for all $\delta$ with $0<|\delta|<\epsilon$, the matrix $M(\delta) = A+\delta B_1+\cdots+\delta^r B_r$ is invertible? Furthermore, is it true that $\|M(\delta)^{-1}\|_{op} \leq C(A, B_1,\cdots,B_r)\delta^{-r}$ for small $\delta$, where $C(\cdots)$ only depends on the fixed matrices?

Answer 1

1(a) and 2(a). Yes. $\det(A+\delta B)$ and $\det(A + \delta B_1 + \ldots + \delta^r B_r)$ are polynomials in $\delta$, and are not constant: the highest order term is $\delta^n \det(B)$ or $\delta^{rn} \det(B_r)$. Therefore there are only finitely many $\delta$ for which this is $0$. For all other $\delta$, including some deleted neighbourhood of $0$, $A + \delta B$ or $A + \delta B_1 + \ldots + \delta^r B_r$ is invertible.

EDIT:

1.(b) and 2(b). No.
Consider $$ A = \pmatrix{0 & 0\cr 0 & 1\cr},\ B = \pmatrix{0 & 1\cr 1 & 0\cr},\ A + \delta B = \pmatrix{0 & \delta\cr \delta & 1\cr}$$ Then $$ (A + \delta B)^{-1} = \pmatrix{-1/\delta^2 & 1/\delta\cr 1/\delta & 0\cr}$$ so $\|(A + \delta B)^{-1} \sim 1/\delta^2$ as $\delta \to 0$.