Let's solve the problem in **two steps** as requested, with explanations after each step. --- ### **Step 1: Should \(A\) be invertible? Why or why not?** #### **Explanation:** - For the system \(Ax = b\) to have a **unique solution**, the matrix \(A\) must be **invertible**. - If \(A\) is **not invertible** (i.e., its determinant is zero), the system either has **no solution** or **infinitely many solutions**. - In the context of the intersection of three planes, invertibility corresponds to the planes meeting at a single unique point. #### **Answer:** Yes, \(A\) **should be invertible** for the system to have a unique solution (a single point of intersection of the three planes). --- ### **Step 2: LU Decomposition by Hand (Construct \(L\) and \(U\))** Given: \[ A = \begin{bmatrix} 7 & 2 & 1 \\ 0 & 3 & -2 \\ -3 & 4 & 2 \end{bmatrix} \] We want to write \(A = LU\), where \(L\) is lower triangular and \(U\) is upper triangular. #### **Procedure:** Let's perform LU decomposition "by hand" (using Doolittle's method, where the diagonal of \(L\) is 1): Let \[ L = \begin{bmatrix} 1 & 0 & 0 \\ l_{21} & 1 & 0 \\ l_{31} & l_{32} & 1 \end{bmatrix} , \quad U = \begin{bmatrix} u_{11} & u_{12} & u_{13} \\ 0 & u_{22} & u_{23} \\ 0 & 0 & u_{33} \end{bmatrix} \] First row of \(U\) is the same as first row of \(A\): \[ u_{11} = 7, \quad u_{12} = 2, \quad u_{13} = 1 \] Now, \(L\) times first row of \(U\) should give first row of \(A\): Done above. Second row: \[ l_{21} \cdot u_{11} = 0 \implies l_{21} = 0 \quad \text{(since second row, first column of \(A\) is 0)} \] Now, \[ u_{22} = 3 \quad \text{(from second row, second column of \(A\): 3)} \] \[ u_{23} = -2 \quad \text{(from second row, third column of \(A\): -2)} \] Third row: \[ l_{31} \cdot u_{11} = -3 \implies l_{31} = \frac{-3}{7} \] \[ l_{31} \cdot u_{12} + l_{32} \cdot u_{22} = 4 \implies \frac{-3}{7} \cdot 2 + l_{32} \cdot 3 = 4 \implies -\frac{6}{7} + 3l_{32} = 4 \implies 3l_{32} = 4 + \frac{6}{7} = \frac{34}{7} \implies l_{32} = \frac{34}{21} \] \[ l_{31} \cdot u_{13} + l_{32} \cdot u_{23} + u_{33} = 2 \implies \frac{-3}{7} \cdot 1 + \frac{34}{21} \cdot (-2) + u_{33} = 2 \implies -\frac{3}{7} - \frac{68}{21} + u_{33} = 2 \] Combine terms: \[ -\frac{3}{7} - \frac{68}{21} = -\frac{9}{21} - \frac{68}{21} = -\frac{77}{21} \] So, \[ -\frac{77}{21} + u_{33} = 2 \implies u_{33} = 2 + \frac{77}{21} = \frac{119}{21} \] #### **Summary:** \[ L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -\frac{3}{7} & \frac{34}{21} & 1 \end{bmatrix} \] \[ U = \begin{bmatrix} 7 & 2 & 1 \\ 0 & 3 & -2 \\ 0 & 0 & \frac{119}{21} \end{bmatrix} \] --- ### **Final Answer (Step 1 & 2):** 1. **A should be invertible** to have a unique solution for the intersection of the three planes. 2. **LU Decomposition** yields: \[ L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -\frac{3}{7} & \frac{34}{21} & 1 \end{bmatrix} , \quad U = \begin{bmatrix} 7 & 2 & 1 \\ 0 & 3 & -2 \\ 0 & 0 & \frac{119}{21} \end{bmatrix} \] where \(A = LU\). --- Let me know if you'd like to continue with the next steps!

Certainly! Continuing from the previous steps, we will now proceed with solving the system using LU decomposition. --- ### **Part 1: Using LU Decomposition to solve \(Ax = b\)** Given: \[ A x = b \] where \[ A = \begin{bmatrix} 7 & 2 & 1 \\ 0 & 3 & -2 \\ -3 & 4 & 2 \end{bmatrix}, \quad b = \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix} \] --- ### **Step 3: Use \(L\) and \(b\) to solve for \(y\) (forward substitution)** **Equation:** \[ L y = b \] With: \[ L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -\frac{3}{7} & \frac{34}{21} & 1 \end{bmatrix} , \quad b = \begin{bmatrix} 2 \\ 4 \\ 6 \end{bmatrix} \] --- **Solve for \(y\):** - \( y_1 \): \[ 1 \cdot y_1 = 2 \implies y_1 = 2 \] - \( y_2 \): \[ 1 \cdot y_2 = 4 \implies y_2 = 4 \] - \( y_3 \): \[ -\frac{3}{7} \cdot y_1 + \frac{34}{21} \cdot y_2 + 1 \cdot y_3 = 6 \] Substitute known \( y_1, y_2 \): \[ -\frac{3}{7} \times 2 + \frac{34}{21} \times 4 + y_3 = 6 \] Calculate: \[ -\frac{6}{7} + \frac{136}{21} + y_3 = 6 \] Express all over denominator 21: \[ -\frac{18}{21} + \frac{136}{21} + y_3 = 6 \] \[ \frac{118}{21} + y_3 = 6 \] \[ y_3 = 6 - \frac{118}{21} = \frac{126}{21} - \frac{118}{21} = \frac{8}{21} \] --- ### **Step 4: Use \(U\) and \(y\) to solve for \(x\) (back substitution)** \[ U x = y \] Where: \[ U = \begin{bmatrix} 7 & 2 & 1 \\ 0 & 3 & -2 \\ 0 & 0 & \frac{119}{21} \end{bmatrix}, \quad y = \begin{bmatrix} 2 \\ 4 \\ \frac{8}{21} \end{bmatrix} \] --- **Solve for \(x_3\):** \[ \frac{119}{21} x_3 = \frac{8}{21} \] \[ x_3 = \frac{\frac{8}{21}}{\frac{119}{21}} = \frac{8/21}{119/21} = \frac{8}{119} \] **Solve for \(x_2\):** \[ 3 x_2 - 2 x_3 = 4 \] \[ 3 x_2 - 2 \times \frac{8}{119} = 4 \] \[ 3 x_2 = 4 + \frac{16}{119} \] Express 4 as \(\frac{476}{119}\): \[ 3 x_2 = \frac{476}{119} + \frac{16}{119} = \frac{492}{119} \] \[ x_2 = \frac{\frac{492}{119}}{3} = \frac{492}{119} \times \frac{1}{3} = \frac{492}{357} = \frac{164}{119} \] **Solve for \(x_1\):** \[ 7 x_1 + 2 x_2 + x_3 = 2 \] Substitute \(x_2\) and \(x_3\): \[ 7 x_1 + 2 \times \frac{164}{119} + \frac{8}{119} = 2 \] Calculate: \[ 7 x_1 + \frac{328}{119} + \frac{8}{119} = 2 \] \[ 7 x_1 + \frac{336}{119} = 2 \] Express 2 as \(\frac{238}{119}\): \[ 7 x_1 = \frac{238}{119} - \frac{336}{119} = -\frac{98}{119} \] \[ x_1 = -\frac{98}{119} \times \frac{1}{7} = -\frac{98}{833} \quad \text{(since } 7 \times 119 = 833) \] Simplify numerator and denominator: \[ x_1 = -\frac{14}{119} = -\frac{2}{17} \] --- ### **### **Final Solution:** \[ \boxed{ x = \begin{bmatrix} -\frac{2}{17} \\ \frac{164}{119} \\ \frac{8}{119} \end{bmatrix} } \] --- ### **Summary:** - The point of intersection of the three planes is: \[ x \approx \left( -0.118, \; 1.38, \; 0.067 \right) \] - **This confirms the system has a unique solution** (since \(A\) is invertible and LU decomposition is successful). --- Let me know if you'd like further explanation or verification!

Certainly! Here's a structured summary of the steps and solutions: --- ### **Part 1: Invertibility of \(A\)** **Question:** Should \(A\) be invertible? Why or why not? **Answer:** Yes, \(A\) should be invertible because only then does the system \(Ax = b\) have a unique solution, corresponding to a single intersection point of the three planes. A non-invertible matrix would imply either no solutions or infinitely many solutions. --- ### **Part 2: LU Decomposition of \(A\)** Given matrix: \[ A = \begin{bmatrix} 7 & 2 & 1 \\ 0 & 3 & -2 \\ -3 & 4 & 2 \end{bmatrix} \] **Construct \(L\) and \(U\):** - \(L\) (lower triangular with 1s on diagonal): \[ L = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ -\frac{3}{7} & \frac{34}{21} & 1 \end{bmatrix} \] - \(U\) (upper triangular): \[ U = \begin{bmatrix} 7 & 2 & 1 \\ 0 & 3 & -2 \\ 0 & 0 & \frac{119}{21} \end{bmatrix} \] **Verification:** Multiplying \(L \times U\) recovers \(A\). --- ### **Part 3: Solving for \(x\) using LU Decomposition** Given \(b = [2, 4, 6]^T\): **Step 1: Forward substitution \(Ly = b\):** \[ \begin{cases} y_1 = 2 \\ y_2 = 4 \\ -\frac{3}{7} \times y_1 + \frac{34}{21} \times y_2 + y_3 = 6 \end{cases} \] Calculate \(y_3\): \[ -\frac{6}{7} + \frac{136}{21} + y_3 = 6 \] Expressed over denominator 21: \[ -\frac{18}{21} + \frac{136}{21} + y_3 = 6 \Rightarrow \frac{118}{21} + y_3 = 6 \Rightarrow y_3 = 6 - \frac{118}{21} = \frac{126}{21} - \frac{118}{21} = \frac{8}{21} \] **Step 2: Back substitution \(Ux = y\):** \[ \begin{cases} \frac{119}{21} x_3 = \frac{8}{21} \Rightarrow x_3 = \frac{8}{119} \\ 3 x_2 - 2 x_3 = 4 \Rightarrow x_2 = \frac{164}{119} \\ 7 x_1 + 2 x_2 + x_3 = 2 \Rightarrow x_1 = -\frac{2}{17} \end{cases} \] --- ### **Final Solution:** \[ \boxed{ x \approx \begin{bmatrix} -\frac{2}{17} \approx -0.118 \\ \frac{164}{119} \approx 1.38 \\ \frac{8}{119} \approx 0.067 \end{bmatrix} } \] This point \((x, y, z)\) is the intersection of the three planes. --- Let me know if you'd like me to clarify any step further!