1. **Problem:** Use the Subspace Test to determine which of the sets are subspaces of $\mathbb{R}^3$. a. All vectors of the form $(a, 0, 0)$. **Step 1:** State the Subspace Test conditions: - The zero vector must be in the set. - The set must be closed under vector addition. - The set must be closed under scalar multiplication. **Step 2:** Check zero vector: - For $a=0$, vector is $(0,0,0)$ which is in the set. **Step 3:** Check closure under addition: - Let $u=(a,0,0)$ and $v=(b,0,0)$ be in the set. - Then $u+v=(a+b,0,0)$ which is in the set. **Step 4:** Check closure under scalar multiplication: - For scalar $c$, $c u = (c a, 0, 0)$ which is in the set. **Conclusion:** The set is a subspace of $\mathbb{R}^3$. --- 19. **Problem:** Determine whether the solution space of $Ax=0$ is a line, plane, or origin only. Find parametric or equation if line or plane. Given matrices: **a.** $A=\begin{bmatrix}1 & -1 & 0 \\ 3 & -1 & -5\end{bmatrix}$ **Step 1:** Write system $Ax=0$: $$\begin{cases} x_1 - x_2 = 0 \\ 3x_1 - x_2 - 5x_3 = 0 \end{cases}$$ **Step 2:** From first equation: $x_1 = x_2$. Substitute into second: $$3x_1 - x_1 - 5x_3 = 0 \implies 2x_1 = 5x_3 \implies x_1 = \frac{5}{2} x_3$$ Since $x_1 = x_2$, we have: $$x_2 = \frac{5}{2} x_3$$ **Step 3:** Parametric form: Let $t = x_3$, then $$x = \left(\frac{5}{2} t, \frac{5}{2} t, t\right) = t \left(\frac{5}{2}, \frac{5}{2}, 1\right)$$ **Conclusion:** Solution space is a line through the origin with parametric equations: $$x_1 = \frac{5}{2} t, \quad x_2 = \frac{5}{2} t, \quad x_3 = t$$ --- 24. **Problem:** Use the Subspace Test to show that - $V = \{p = a_0 + a_1 x + a_2 x^2 + a_3 x^3 : a_0 + a_3 = 0\}$ - $W = \{p : p'(1) = 0\}$ are subspaces of $P_3$. **Step 1:** Check $V$: - Zero polynomial has $a_0=0$, $a_3=0$, so $a_0 + a_3=0$. - Closure under addition: If $p, q \in V$, then $(a_0 + a_3) + (b_0 + b_3) = 0 + 0 = 0$. - Closure under scalar multiplication: For scalar $c$, $c a_0 + c a_3 = c(a_0 + a_3) = c \cdot 0 = 0$. **Step 2:** Check $W$: - $p'(x) = a_1 + 2 a_2 x + 3 a_3 x^2$. - Condition $p'(1) = a_1 + 2 a_2 + 3 a_3 = 0$. - Zero polynomial derivative is zero, so in $W$. - Closure under addition: If $p, q \in W$, then $p'(1) + q'(1) = 0 + 0 = 0$. - Closure under scalar multiplication: For scalar $c$, $(c p)'(1) = c p'(1) = c \cdot 0 = 0$. **Step 3:** Intersection $V \cap W$: - Polynomials satisfying both $a_0 + a_3 = 0$ and $a_1 + 2 a_2 + 3 a_3 = 0$. - Zero polynomial satisfies both. - Closure under addition and scalar multiplication hold as both conditions are linear. **Conclusion:** $V$, $W$, and $V \cap W$ are subspaces of $P_3$.