Suppose $A \in \mathcal M(n \times m; \mathbb R)$ is fixed with $n > m$. Let $f: [0,1] \to \mathcal M(n \times n; \mathbb R)$ be a continuous function. Let $X_1$ and $X_2$ be fixed matrices in $\mathcal M(m \times n; \mathbb R)$ such that $f(0)=AX_1$ and $f(1) = AX_2$. If we suppose for every $t \in [0,1]$, there exists at least one $X \in \mathcal M(n \times n; \mathbb R)$such that $AX=f(t)$. We know there is at least one solution trajectory $X(t) \mapsto AX(t)=f(t)$. Is it possible to choose this solution path continuously?
My idea: Let $A^{\dagger}$ be the Moore–Penrose Generalized Inverse and it is uniquely defined. Then a solution to the linear system $AX=f(t)$ (since we are assuming solvability) has the form $X = A^{\dagger} f(t) + (I - AA^{\dagger})z$ for all $z \in \mathbb R$. Now Let $z_1, z_2$ be such that $AX_1 = A^{\dagger} f(0) + (I-AA^{\dagger})z_1$ and $AX_1 = A^{\dagger} f(1) + (I-AA^{\dagger})z_2$. Let $X(t) = A^{\dagger} f(t) + (I - AA^{\dagger})((1-t)z_1 + tz_2)$. This should be a continuous solution path.