1. **Stating the problem:** We are given two groups: (i) $\langle M_2^*, \times \rangle$ is the group of all $2 \times 2$ real matrices with nonzero determinant under matrix multiplication. (ii) $\langle \mathbb{R}^*, \times \rangle$ is the group of all nonzero real numbers under multiplication. A map $\gamma : M_2^* \to \mathbb{R}^*$ is defined by $\gamma(M) = \det(M)$ for each $M \in M_2^*$. We need to investigate whether $\gamma$ is a group homomorphism. 2. **Recall the definition of a group homomorphism:** A map $\phi : G \to H$ between groups $(G, \cdot)$ and $(H, *)$ is a homomorphism if for all $a,b \in G$, $$\phi(a \cdot b) = \phi(a) * \phi(b).$$ 3. **Apply the definition to $\gamma$:** We want to check if for all $A,B \in M_2^*$, $$\gamma(A \times B) = \gamma(A) \times \gamma(B).$$ 4. **Use the property of determinants:** It is a well-known property that for square matrices, $$\det(A \times B) = \det(A) \times \det(B).$$ 5. **Check the condition:** Substitute into the homomorphism condition: $$\gamma(A \times B) = \det(A \times B) = \det(A) \times \det(B) = \gamma(A) \times \gamma(B).$$ 6. **Conclusion:** Since the condition holds for all $A,B \in M_2^*$, the map $\gamma$ preserves the group operation. Therefore, $\gamma$ is a group homomorphism from $\langle M_2^*, \times \rangle$ to $\langle \mathbb{R}^*, \times \rangle$. **Final answer:** $$\boxed{\gamma \text{ is a group homomorphism.}}$$