Let's solve **Problem 23** as an example, following your instructions for two steps with explanations at each step, then provide the final answer. --- ## Problem 23 Given: - \(\mathbf{u} = (2, , 3)\) - \(\mathbf{v} = (5, 4, -2)\) - \(\mathbf{w} = (2, -1, 1)\) Determine if they are linearly dependent or independent. --- ### **Step 1: Set up the linear dependence equation** We want to find scalars \(a\), \(b\), and \(c\) (not all zero) such that: \[ a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = (, , ) \] That is: \[ a(2,,3) + b(5,4,-2) + c(2,-1,1) = (,,) \] Expanding this: - x-component: \(2a + 5b + 2c = \) - y-component: \(a + 4b - 1c = \) - z-component: \(3a - 2b + 1c = \) So the system is: \[ \begin{cases} 2a + 5b + 2c = \\ 4b - c = \\ 3a - 2b + c = \end{cases} \] **Explanation:** We wrote the vector equation and expanded it into a system of linear equations for the coefficients. --- ### **Step 2: Solve the system** From the second equation: \(4b - c = \implies c = 4b\). Substitute \(c = 4b\) into the other equations: 1. \(2a + 5b + 2c = \) \[ 2a + 5b + 2(4b) = \implies 2a + 5b + 8b = \implies 2a + 13b = \implies a = -\frac{13}{2}b \] 2. \(3a - 2b + c = \) \[ 3a - 2b + 4b = \implies 3a + 2b = \] Substitute \(a\) from above: \[ 3(-\frac{13}{2}b) + 2b = \implies -\frac{39}{2}b + 2b = \implies -\frac{39}{2}b + \frac{4}{2}b = \implies -\frac{35}{2}b = \] So \(b = \). If \(b = \), then \(c = \), and \(a = \). **Explanation:** We substituted and simplified to discover that all coefficients must be zero. --- ## **Final Answer** **The only solution is \(a = b = c = \), so the vectors are linearly independent.** --- Let me know if you'd like a solution for another problem in the same format!

# Problem Solutions for Problems 26, 30, 31, and 32 ## Problem 26 Given the vectors: - \(\mathbf{u} = (1, -2, 2)\) - \(\mathbf{v} = (0, 1, 3)\) - \(\mathbf{w} = (1, -1, 4)\) ### **Step 1: Set up the linear dependence equation** We want to find scalars \(a\), \(b\), and \(c\) such that: \[ a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = (0, 0, 0) \] Expanding this, we get: - x-component: \(a + c = 0\) - y-component: \(-2a + b - c = 0\) - z-component: \(2a + 3b + 4c = 0\) The system of equations becomes: \[ \begin{cases} a + c = 0 \\ -2a + b - c = 0 \\ 2a + 3b + 4c = 0 \end{cases} \] **Explanation:** We expanded the equation and formed a system of linear equations based on the components of the vectors. ### **Step 2: Solve the system** From the first equation, we have: \[ c = -a \] Substituting \(c = -a\) into the second equation: \[ -2a + b - (-a) = 0 \implies -2a + b + a = 0 \implies b - a = 0 \implies b = a \] Now substitute \(b = a\) and \(c = -a\) into the third equation: \[ 2a + 3(a) + 4(-a) = 0 \implies 2a + 3a - 4a = 0 \implies a = 0 \] Thus, \(b = 0\) and \(c = 0\). **Explanation:** All coefficients must be zero, indicating the vectors are linearly independent. ## **Final Answer for Problem 26** The vectors \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are linearly independent. --- ## Problem 30 Given the vectors: - \(\mathbf{u} = (2, 1, 3)\) - \(\mathbf{v} = (4, 2, 6)\) - \(\mathbf{w} = (1, 0, 1)\) ### **Step 1: Set up the linear dependence equation** We want to determine if there exist scalars \(a\), \(b\), and \(c\) such that: \[ a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = (0, 0, 0) \] Expanding this gives: - x-component: \(2a + 4b + c = 0\) - y-component: \(a + 2b = 0\) - z-component: \(3a + 6b + c = 0\) The system of equations is: \[ \begin{cases} 2a + 4b + c = 0 \\ a + 2b = 0 \\ 3a + 6b + c = 0 \end{cases} \] **Explanation:** We set up the equations based on the vector components to check for linear dependence. ### **Step 2: Solve the system** From the second equation: \[ a = -2b \] Substituting \(a = -2b\) into the first equation: \[ 2(-2b) + 4b + c = 0 \implies -4b + 4b + c = 0 \implies c = 0 \] Now substituting \(c = 0\) into the third equation: \[ 3(-2b) + 6b + 0 = 0 \implies -6b + 6b = 0 \] This holds true for any \(b\). Thus, choose \(b = 1\), then \(a = -2\) and \(c = 0\). **Explanation:** We found non-zero solutions for the coefficients, indicating dependence. ## **Final Answer for Problem 30** The vectors \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are linearly dependent, with scalars \(a = -2\), \(b = 1\), and \(c = 0\). --- ## Problem 31 Given the vectors: - \(\mathbf{u} = (1, 2, 3)\) - \(\mathbf{v} = (2, 4, 6)\) - \(\mathbf{w} = (3, 1, 2)\) ### **Step 1: Set up the linear dependence equation** We want to find scalars \(a\), \(b\), and \(c\) such that: \[ a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = (0, 0, 0) \] Expanding this gives: - x-component: \(a + 2b + 3c = 0\) - y-component: \(2a + 4b + c = 0\) - z-component: \(3a + 6b + 2c = 0\) The system of equations is: \[ \begin{cases} a + 2b + 3c = 0 \\ 2a + 4b + c = 0 \\ 3a + 6b + 2c = 0 \end{cases} \] **Explanation:** We formed a system of equations from the vector components to investigate linear dependence. ### **Step 2: Solve the system** From the second equation: \[ 2a + 4b + c = 0 \implies c = -2a - 4b \] Substituting \(c\) into the first equation: \[ a + 2b + 3(-2a - 4b) = 0 \implies a + 2b - 6a - 12b = 0 \implies -5a - 10b = 0 \implies a + 2b = 0 \implies a = -2b \] Now substituting back into the equation for \(c\): \[ c = -2(-2b) - 4b = 4b - 4b = 0 \] Setting \(b = 1\), we have \(a = -2\) and \(c = 0\). **Explanation:** We found non-zero solutions for the coefficients, indicating dependence. ## **Final Answer for Problem 31** The vectors \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are linearly dependent, with scalars \(a = -2\), \(b = 1\), and \(c = 0\). --- ## Problem 32 Given the vectors: - \(\mathbf{u} = (0, 1, 2)\) - \(\mathbf{v} = (1, 2, 3)\) - \(\mathbf{w} = (2, 3, 4)\) ### **Step 1: Set up the linear dependence equation** We want to find scalars \(a\), \(b\), and \(c\) such that: \[ a\mathbf{u} + b\mathbf{v} + c\mathbf{w} = (0, 0, 0) \] Expanding this gives: - x-component: \(b + 2c = 0\) - y-component: \(a + 2b + 3c = 0\) - z-component: \(2a + 3b + 4c = 0\) The system is: \[ \begin{cases} b + 2c = 0 \\ a + 2b + 3c = 0 \\ 2a + 3b + 4c = 0 \end{cases} \] **Explanation:** We established the system of equations based on the vector components. ### **Step 2: Solve the system** From the first equation: \[ b = -2c \] Substituting \(b = -2c\) into the second equation: \[ a + 2(-2c) + 3c = 0 \implies a - 4c + 3c = 0 \implies a - c = 0 \implies a = c \] Substituting \(a = c\) and \(b = -2c\) into the third equation: \[ 2c + 3(-2c) + 4c = 0 \implies 2c - 6c + 4c = 0 \implies 0 = 0 \] This holds for any \(c\). Choose \(c = 1\), then \(a = 1\) and \(b = -2\). **Explanation:** We derived non-zero coefficients for the vectors, indicating dependence. ## **Final Answer for Problem 32** The vectors \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) are linearly dependent, with scalars \(a = 1\), \(b = -2\), and \(c = 1\).