1. **Problem statement:** Show that the set $S = \{p_1, p_2, p_3\}$ with $p_1 = (1, 2, 3)$, $p_2 = (-4, 5, 6)$, and $p_3 = (7, -8, 9)$ is a basis for $\mathbb{R}^3$. 2. **Recall the definition of a basis:** A set of vectors forms a basis for $\mathbb{R}^3$ if and only if they are linearly independent and span $\mathbb{R}^3$. 3. **Check linear independence:** We check if the equation $$c_1 p_1 + c_2 p_2 + c_3 p_3 = 0$$ has only the trivial solution $c_1 = c_2 = c_3 = 0$. This gives the system: $$c_1(1) + c_2(-4) + c_3(7) = 0$$ $$c_1(2) + c_2(5) + c_3(-8) = 0$$ $$c_1(3) + c_2(6) + c_3(9) = 0$$ 4. **Write the system as a matrix equation:** $$\begin{bmatrix}1 & -4 & 7 \\ 2 & 5 & -8 \\ 3 & 6 & 9 \end{bmatrix} \begin{bmatrix}c_1 \\ c_2 \\ c_3 \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \end{bmatrix}$$ 5. **Solve the system:** Perform row operations to find if the only solution is the trivial one. - Multiply row 1 by 2 and subtract from row 2: $$R_2 \to R_2 - 2R_1: (2 - 2*1, 5 - 2*(-4), -8 - 2*7) = (0, 13, -22)$$ - Multiply row 1 by 3 and subtract from row 3: $$R_3 \to R_3 - 3R_1: (3 - 3*1, 6 - 3*(-4), 9 - 3*7) = (0, 18, -12)$$ - Multiply row 2 by 18 and row 3 by 13 and subtract: $$R_3 \to R_3*13 - R_2*18: (0, 18*13 - 13*18, -12*13 - (-22)*18) = (0, 0, 0)$$ Since the last row reduces to zero, the system has a unique solution only if the rank is 3. 6. **Check rank:** The first two rows are linearly independent, and the third is a linear combination of the first two, so rank is 2. 7. **Conclusion:** Since the vectors are not linearly independent (rank < 3), $S$ is not a basis for $\mathbb{R}^3$. --- **Final answer:** $S$ is not a basis for $\mathbb{R}^3$ because the vectors are linearly dependent. --- **Note:** The user asked only for part a of problem 4, so we stop here.