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

To perform LU decomposition using Doolittle's algorithm, we set the diagonal elements of the lower triangular matrix $L$ to $1$, so $L_{11} = L_{22} = L_{33} = 1$.

The coefficient matrix $A$ is:

Applying Gaussian elimination to find $U$ and the multipliers for $L$:

1. First column elimination:

   • Row 2: $R_2 \leftarrow R_2 - (1) R_1 \implies L_{21} = 1$, giving $\begin{pmatrix} 0 & 2 & 1 \end{pmatrix}$.

   • Row 3: $R_3 \leftarrow R_3 - (2) R_1 \implies L_{31} = 2$, giving $\begin{pmatrix} 0 & -1 & -1 \end{pmatrix}$.

2. Second column elimination:

   • Row 3: $R_3 \leftarrow R_3 - \left(-\frac{1}{2}\right) R_2 \implies L_{32} = -\frac{1}{2}$.

   • The resulting third row element is:

     $$U_{33} = -1 - \left(-\frac{1}{2}\right)(1) = -1 + \frac{1}{2} = -\frac{1}{2}$$

3. Solving the system of equations $Ax = b$: $$\begin{aligned} x_1 + x_2 - 2x_3 &= 4 \\ 2x_2 + x_3 &= 7 - 4 = 3 \\ -x_2 - x_3 &= 7 - 2(4) = -1 \implies x_2 + x_3 = 1 \end{aligned}$$

   From $2x_2 + x_3 = 3$ and $x_2 + x_3 = 1$:

   $$x_2 = 2, \quad x_3 = -1$$

   Substitute $x_2$ and $x_3$ into the first equation:

   $$x_1 + 2 - 2(-1) = 4 \implies x_1 + 4 = 4 \implies x_1 = 0$$

Thus, $L_{32} = -\frac{1}{2}$, $U_{33} = -\frac{1}{2}$, and $x_1 = 0$, which corresponds to option (D).