1. **Problem statement:** Determine which subsets of $\text{Mat}_2(\mathbb{R})$ are subspaces. 2. **Recall:** A subset $W$ of a vector space $V$ is a subspace if it satisfies: - The zero vector is in $W$. - $W$ is closed under vector addition. - $W$ is closed under scalar multiplication. 3. **Check $M_1 = \{ A \in \text{Mat}_2(\mathbb{R}) : \text{rank}(A) = 2 \}$:** - The zero matrix has rank 0, not 2, so $0 \notin M_1$. - Hence, $M_1$ is not a subspace. 4. **Check $M_2 = \left\{ \begin{bmatrix} a & b \\ c & d \end{bmatrix} : a + \alpha d = 0 \right\}$:** - The zero matrix satisfies $0 + \alpha \cdot 0 = 0$, so $0 \in M_2$. - For $A, B \in M_2$, with $a_A + \alpha d_A = 0$ and $a_B + \alpha d_B = 0$, then for any scalar $k$: $$ (a_A + a_B) + \alpha (d_A + d_B) = (a_A + \alpha d_A) + (a_B + \alpha d_B) = 0 + 0 = 0 $$ $$ k a_A + \alpha k d_A = k(a_A + \alpha d_A) = k \cdot 0 = 0 $$ - So $M_2$ is closed under addition and scalar multiplication. 5. **Conclusion:** - $M_1$ is not a subspace. - $M_2$ is a subspace of $\text{Mat}_2(\mathbb{R})$. **Final answer:** $M_2$ is a subspace; $M_1$ is not.