Let $M(t)\in \mathbb{R}^{n\times n}$ be a continuous mapping over $[0,T]$. And let us assume that $M(t)$ symmetric and positive-definite matrix for all $t\in [0,T]$.
Can we write the following :
$$\Big(\int_a^b M(t)dt\Big)^{-1}= \int_a^b M^{-1} (t)dt,$$ for every $a,b \in [0,T]$ ?
On
This equality will simply not satisfy when the length $(b-a)$ and $det(M)(t)$ do not observe each other.
On
It's probably hard to find cases where the equality does hold. You already have a counterexample if you take $M(t)=I _n$. Then $$ \Big(\int_a^bM(t)\,dt\Big)^{-1} =\frac1{b-a}\,I_n, $$ while $$ \int_a^bM(t)^{-1}\,dt=(b-a)\,I_n. $$ So even with this trivial of trivials example the equality will fail as long as $b-a\ne\pm1$.
No. Choose $$ M(t) = \begin{pmatrix} \exp(t) & 0 \\ 0 & \exp(t) \end{pmatrix} \implies M(t)^{-1} = \begin{pmatrix} \exp(-t) & 0 \\ 0 & \exp(-t) \end{pmatrix}. $$ Then $$ \left( \int^b_a M(t)~\mathrm{d}t\right)^{-1} = \begin{pmatrix} \frac{1}{\exp(b)-\exp(a)} & 0 \\ 0 & \frac{1}{\exp(b)-\exp(a)} \end{pmatrix} $$ and $$ \int^b_a M(t)^{-1}~\mathrm{d}t = \begin{pmatrix} -\exp(-b)+\exp(-a)) & 0 \\ 0 & -\exp(-b)+\exp(a) \end{pmatrix}. $$ Different story though if inversion was a linear operator of matrices...