Question

Consider solving the following system of simultaneous equations using LU decomposition.  \[ x_1 + x_2 - 2x_3 = 4 \\ x_1 + 3x_2 - x_3 = 7 \\ 2x_1 + x_2 - 5x_3 = 7 \] where \( L \) and \( U \) are denoted as \[ L = \begin{pmatrix} L_{11} & 0 & 0 \\ L_{21} & L_{22} & 0 \\ L_{31} & L_{32} & L_{33} \end{pmatrix}, U = \begin{pmatrix} U_{11} & U_{12} & U_{13} \\ 0 & U_{22} & U_{23} \\ 0 & 0 & U_{33} \end{pmatrix} \] Which one of the following is the correct combination of values for \( L_{32}, U_{33}, \) and \( x_1 \)?

• A. \( L_{32} = 2, U_{33} = -\frac{1}{2}, x_1 = -1 \)
• B. \( L_{32} = 2, U_{33} = 2, x_1 = -1 \)
• C. \( L_{32} = -\frac{1}{2}, U_{33} = 2, x_1 = 0 \)
• D. \( L_{32} = -\frac{1}{2}, U_{33} = -\frac{1}{2}, x_1 = 0 \)

Hint:

In LU decomposition, remember that the lower triangular matrix \( L \) has ones on its diagonal, and the upper triangular matrix \( U \) has all zero entries below the diagonal.

Answer 1

The Correct Option is D

To solve this problem, we apply LU decomposition to decompose the given system into lower triangular matrix \( L \) and upper triangular matrix \( U \). The system can be written as: \[ A \cdot \mathbf{x} = \mathbf{b} \] where \[ A = \begin{pmatrix} 1 & 1 & -2 \\ 1 & 3 & -1 \\ 2 & 1 & -5 \end{pmatrix}, \mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \mathbf{b} = \begin{pmatrix} 4 \\ 7 \\ 7 \end{pmatrix} \] Step 1: LU Decomposition Performing LU decomposition involves expressing \( A \) as the product of a lower triangular matrix \( L \) and an upper triangular matrix \( U \). We begin by setting up the equations for \( L \) and \( U \):
1. From the first row: \( U_{11} = 1 \), \( U_{12} = 1 \), \( U_{13} = -2 \)
2. From the second row: \( L_{21} = 1 \), \( L_{22} = 2 \), \( U_{23} = -1 \) 3. From the third row: \( L_{31} = 2 \), \( L_{32} = -\frac{1}{2} \), \( U_{33} = -\frac{1}{2} \)

Step 2: Solve for \( x_1 \) By forward substitution and solving for \( x_1, x_2, x_3 \), we obtain \( x_1 = 0 \), confirming that the value of \( x_1 \) is 0.

Final Answer Thus, the correct values are: \[ L_{32} = -\frac{1}{2}, U_{33} = -\frac{1}{2}, x_1 = 0. \] Final Answer: (D)