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No, this is trivially false. Start with $\mathcal{A}, \mathcal{B} \subset B(H)$ that are not a counterexample and define $$\mathcal{A}^{(2)} = \{A \oplus A \in B(H \oplus H): A \in \mathcal{A}\}$$ and $$\mathcal{B}^{(2)} = \{B \oplus B \in B(H \oplus H): B \in \mathcal{B}\}.$$ They and their commutants are still $II_1$ factors and their intersection is still trivial, but their commutants $M_2(\mathcal{A})$ and $M_2(\mathcal{B})$ now intersect in the scalar matrices.

No, this is trivially false. Start with $\mathcal{A}, \mathcal{B} \subset B(H)$ that are not a counterexample and define $$\mathcal{A}^{(2)} = \{A \oplus A \in B(H \oplus H): A \in \mathcal{A}\}$$ and $$\mathcal{B}^{(2)} = \{B \oplus B \in B(H \oplus H): B \in \mathcal{B}\}.$$ They and their commutants are still $II_1$ factors and their intersection is still trivial, but their commutants $M_2(\mathcal{A}')$ and $M_2(\mathcal{B}')$ now intersect in the scalar matrices. The answer is now in its true, correct, and final form. I will notify the Cyber Police of any additional unauthorized edits.

No, this is trivially false. Start with $\mathcal{A}, \mathcal{B} \subset B(H)$ that are not a counterexample and define $$\mathcal{A}^{(2)} = \{A \oplus A \in B(H \oplus H): A \in \mathcal{A}\}$$ and $$\mathcal{B}^{(2)} = \{B \oplus B \in B(H \oplus H): B \in \mathcal{B}\}.$$ They and their commutants are still $II_1$ factors and their intersection is still trivial, but their commutants $M_2(\mathcal{A}')$ and $M_2(\mathcal{B}')$ don't intersect in the scalar matrices.

No, this is trivially false. Start with $\mathcal{A}, \mathcal{B} \subset B(H)$ that are not a counterexample and define $$\mathcal{A}^{(2)} = \{A \oplus A \in B(H \oplus H): A \in \mathcal{A}\}$$ and $$\mathcal{B}^{(2)} = \{B \oplus B \in B(H \oplus H): B \in \mathcal{B}\}.$$ They and their commutants are still $II_1$ factors and their intersection is still trivial, but their commutants $M_2(\mathcal{A})$ and $M_2(\mathcal{B})$ now intersect in the scalar matrices.

No, this is trivially false. Start with $\mathcal{A}, \mathcal{B} \subset B(H)$ that are not a counterexample and define $$\mathcal{A}^{(2)} = \{A \oplus A \in B(H \oplus H): A \in \mathcal{A}\}$$ and $$\mathcal{B}^{(2)} = \{B \oplus B \in B(H \oplus H): B \in \mathcal{B}\}.$$ They and their commutants are still $II_1$ factors and their intersection is still trivial, but their commutants $M_2(\mathcal{A})$ and $M_2(\mathcal{B})$ now intersect in the scalar matrices.

No, this is trivially false. Start with $\mathcal{A}, \mathcal{B} \subset B(H)$ that are not a counterexample and define $$\mathcal{A}^{(2)} = \{A \oplus A \in B(H \oplus H): A \in \mathcal{A}\}$$ and $$\mathcal{B}^{(2)} = \{B \oplus B \in B(H \oplus H): B \in \mathcal{B}\}.$$ They and their commutants are still $II_1$ factors and their intersection is still trivial, but their commutants $M_2(\mathcal{A}')$ and $M_2(\mathcal{B}')$ don't intersect in the scalar matrices.