1. **Stating the problem:** Given vectors \( u = (2,4,-5) \) and \( v = (1,-6,9) \), perform the following vector operations: (a) Calculate \( u + v \), \( 7u \), \( -v \), and \( 3u - 5v \). (b) Explain the zero vector \( 0 \) in \( \mathbb{R}^n \) and its property with vector addition. (c) Given \( u = \begin{bmatrix} 2 \\ 3 \\ -4 \end{bmatrix} \) and \( v = \begin{bmatrix} 3 \\ -1 \\ -2 \end{bmatrix} \), compute \( 2u - 3v \). 2. **Formulas and rules:** - Vector addition: \( u + v = (u_1 + v_1, u_2 + v_2, ..., u_n + v_n) \). - Scalar multiplication: \( a u = (a u_1, a u_2, ..., a u_n) \). - Zero vector \( 0 = (0,0,...,0) \) satisfies \( u + 0 = u \). 3. **Step-by-step solutions:** **(a)** - \( u + v = (2+1, 4+(-6), -5+9) = (3, -2, 4) \). - \( 7u = (7 \times 2, 7 \times 4, 7 \times -5) = (14, 28, -35) \). - \( -v = (-1) \times (1, -6, 9) = (-1, 6, -9) \). - \( 3u - 5v = 3u + (-5v) = (3 \times 2, 3 \times 4, 3 \times -5) + (-5 \times 1, -5 \times -6, -5 \times 9) = (6, 12, -15) + (-5, 30, -45) = (1, 42, -60) \). **(b)** - The zero vector \( 0 = (0,0,...,0) \) acts like the number zero in scalar arithmetic. - For any vector \( u = (a_1, a_2, ..., a_n) \), adding zero vector gives: \( u + 0 = (a_1 + 0, a_2 + 0, ..., a_n + 0) = (a_1, a_2, ..., a_n) = u \). **(c)** - Compute \( 2u = 2 \times \begin{bmatrix} 2 \\ 3 \\ -4 \end{bmatrix} = \begin{bmatrix} 4 \\ 6 \\ -8 \end{bmatrix} \). - Compute \( 3v = 3 \times \begin{bmatrix} 3 \\ -1 \\ -2 \end{bmatrix} = \begin{bmatrix} 9 \\ -3 \\ -6 \end{bmatrix} \). - Then \( 2u - 3v = \begin{bmatrix} 4 \\ 6 \\ -8 \end{bmatrix} - \begin{bmatrix} 9 \\ -3 \\ -6 \end{bmatrix} = \begin{bmatrix} 4 - 9 \\ 6 - (-3) \\ -8 - (-6) \end{bmatrix} = \begin{bmatrix} -5 \\ 9 \\ -2 \end{bmatrix} \). 4. **Final answers:** - (a) \( u + v = (3, -2, 4) \), \( 7u = (14, 28, -35) \), \( -v = (-1, 6, -9) \), \( 3u - 5v = (1, 42, -60) \). - (b) The zero vector \( 0 \) satisfies \( u + 0 = u \). - (c) \( 2u - 3v = \begin{bmatrix} -5 \\ 9 \\ -2 \end{bmatrix} \).