Question 45

The coefficient matrix is:

Using Doolittle's LU decomposition with unit diagonal for $L$ ($L_{11} = L_{22} = L_{33} = 1$):

1. First row of $U$:

   $$U_{11} = A_{11} = 1, \quad U_{12} = A_{12} = 1, \quad U_{13} = A_{13} = -2$$

2. First column of $L$:

   $$L_{21} = \frac{A_{21}}{U_{11}} = \frac{1}{1} = 1, \quad L_{31} = \frac{A_{31}}{U_{11}} = \frac{2}{1} = 2$$

3. Second row of $U$:

   $$U_{22} = A_{22} - L_{21} U_{12} = 3 - 1(1) = 2$$
   $$U_{23} = A_{23} - L_{21} U_{13} = -1 - 1(-2) = 1$$

4. Second column of $L$:

   $$L_{32} = \frac{A_{32} - L_{31} U_{12}}{U_{22}} = \frac{1 - 2(1)}{2} = -\frac{1}{2}$$

5. Third row of $U$:

   $$U_{33} = A_{33} - L_{31} U_{13} - L_{32} U_{23} = -5 - 2(-2) - \left(-\frac{1}{2}\right)(1) = -5 + 4 + \frac{1}{2} = -\frac{1}{2}$$

Now solve the system of linear equations:

1. $x_1 + x_2 - 2x_3 = 4$
2. $x_1 + 3x_2 - x_3 = 7$
3. $2x_1 + x_2 - 5x_3 = 7$

Subtracting (1) from (2): $2x_2 + x_3 = 3 \implies x_3 = 3 - 2x_2$.
Substituting into (1): $x_1 + x_2 - 2(3 - 2x_2) = 4 \implies x_1 + 5x_2 = 10$.
Substituting into (3): $2x_1 + x_2 - 5(3 - 2x_2) = 7 \implies 2x_1 + 11x_2 = 22$.
Multiplying $x_1 + 5x_2 = 10$ by 2 gives $2x_1 + 10x_2 = 20$. Subtracting yields $x_2 = 2$.
Then $x_1 = 10 - 5(2) = 0$.

Thus, $L_{32} = -\frac{1}{2}$, $U_{33} = -\frac{1}{2}$, and $x_1 = 0$, matching Option (D).