 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2010/p/1/solution

Eliminate the first column with multipliers

ℓ_{21} = 2

and

ℓ_{31} = - 1

. The new second and third rows are

(0, 2, 4)

and

(0, - 1, 3)

. The next multiplier is

ℓ_{32} = - 1/2

, giving final row

(0, 0, 5)

. Recording these multipliers gives the LU decomposition

L = 12 - 1 01 - 1/2 001, U = 100220245, A = LU .

For an upper-triangular matrix with nonzero diagonal, the back substitution algorithm works from bottom to top:

x_{i} = \frac{b _{i} - \sum _{j = i + 1}^{n} u _{ij} x _{j}}{u _{ii}}, i = n, n - 1, \dots, 1.

In the separate triangular system printed in the question,

5 x_{3} = 10

2 x_{2} + 4 x_{3} = 6

, and

x_{1} + 2 x_{2} + 2 x_{3} = 3

. Hence

x = (1, - 1, 2)^{T} .

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