I have a matrix that is a linear combination of symmetric square matrices:
$$A_t = \sum_{j = 1}^{i - 1} (t_j - t_{j - 1}) B_j + (t - t_{i - 1}) B_i$$
Here $t_j > t_{j - 1}$ and $t_{i - 1} \le t \le t_i$ for some $i \le M + 1$.
The matrix $B_1$ is positive definite and the rest are only assumed to be positive semidefinite. Additionally, there are only ever $M + 1$ matrices at most, so we can bound $\left\|B_i\right\| \le C$ for some $C > 0$.
There is a sense in which we can say that $A_t = O(t)$, in that it's easy to show that $\left\| A_t \right\| \le Ct$. But now I want to say that $A_t^{-1} = O(1/t)$ in a similar way. For example, if $A_t$ was just $t B_1$ this would be easy as $A_t^{-1} = \frac{1}{t} B_1^{-1}$. But when summing the matrices this becomes much harder to do as I can't seem to get a good upper bound for inverse matrices.
If I'm too vague let me know and I can add more detail, but I appreciate any assistance.
This is not true. For instance, if $B_1=I_2$ and $B_i=\operatorname{diag}(1,0)$ for every $i\ge2$, then $A_t=\operatorname{diag}(t,1)$ when $t\ge1$. It follows that $A_t=O(t)$ when $t\to+\infty$ but $\|A_t^{-1}\|_2=1$ for every $t\ge1$.