Let $A\in L^\infty(\Omega;\Bbb R^{2\times 2})$, be positive definite such that $A^\dagger = A$, where $A^\dagger$ is the pseudo-inverse of $A$, one of the attributes of being a pseudo-inverse is that $AA^\dagger A=A$, assuming $A\ne 0,A\ne I_d$, then $A = A^3$ and so $A=\lim_{n\to\infty}A^{3n}=0$ if the largest eigenvalue of $A$ is less than $1$, or $\infty$ if the largest eigenvalue is greater than $1$.
Does this imply in fact that $A=A^\dagger=A^{-1}$, since if this is true then nothing "blows up"?
I am stricty speaking for matrices because I am not really sure what $L^\infty (\Omega; \mathbb{R}^{2\times 2}) $ means in your context.
For a matrix $A\in \mathbb{R}^{n \times n}$ we have $A^\dagger = A^{-1}$ iff $A$ is invertible. Assuming $A$ is not invertible. Then we have $x\neq 0$ with $Ax=0$. This yields $x^T A x = 0$. Therefore $A$ can not be positive definite.