1. **Problem Statement:** Check whether the following sets form subspaces of their respective vector spaces. 2. **Recall the Subspace Criteria:** A subset $S$ of a vector space $V$ is a subspace if and only if: - The zero vector of $V$ is in $S$. - $S$ is closed under vector addition. - $S$ is closed under scalar multiplication. --- ### (i) $S = \{(x_1, x_2, x_3) \in \mathbb{R}^3 \mid x_1 - 4x_2 + 5x_3 = 2\}$ 3. **Check if zero vector is in $S$:** The zero vector in $\mathbb{R}^3$ is $(0,0,0)$. Substitute into the equation: $$0 - 4\cdot0 + 5\cdot0 = 0 \neq 2$$ So, zero vector is **not** in $S$. 4. **Conclusion:** Since zero vector is not in $S$, $S$ is **not** a subspace of $\mathbb{R}^3$. --- ### (ii) $S = \{(x_1, x_2) \in \mathbb{R}^2 \mid x_2 = x_1^2\}$ 5. **Check zero vector:** Zero vector is $(0,0)$. Check if it satisfies $x_2 = x_1^2$: $$0 = 0^2 = 0$$ Zero vector is in $S$. 6. **Check closure under addition:** Take two vectors in $S$: $$u = (a, a^2), \quad v = (b, b^2)$$ Sum: $$u + v = (a+b, a^2 + b^2)$$ For $u+v$ to be in $S$, second component must equal square of first: $$a^2 + b^2 \stackrel{?}{=} (a+b)^2 = a^2 + 2ab + b^2$$ This is false unless $ab=0$. So, $S$ is **not** closed under addition. 7. **Conclusion:** $S$ is **not** a subspace of $\mathbb{R}^2$. --- ### (iii) $S = \{A \in M_{2\times2} \mid \det(A) = 0\}$ 8. **Check zero matrix:** Zero matrix $0$ has determinant 0, so $0 \in S$. 9. **Check closure under addition:** Take two singular matrices $A, B$ with $\det(A) = 0$ and $\det(B) = 0$. Sum $A+B$ may have nonzero determinant. Example: $$A = \begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix}, B = \begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$$ Both have determinant 0, but $$A+B = \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}$$ which has determinant 1. So, $S$ is **not** closed under addition. 10. **Conclusion:** $S$ is **not** a subspace of $M_{2\times2}$. --- **Final answers:** - (i) Not a subspace - (ii) Not a subspace - (iii) Not a subspace