 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/digital-systems-linear-algebra-introduction-to-formal-proof/6/a/ii/solution

From zero, the first update is $x_{1} = (- 1, 3/2, 1/4)^{T}$ . For the next update, the first component is

- 1 + 3/2 - 1/4 = 1/4

, the second is

3/2 - (1/2) (3/2) = 3/4

, and the third is

1/4 - (3/4) (3/2) + (1/2) (1/4) = - 3/4

. Therefore $x_{2} = (1/4, 3/4, - 3/4)^{T}$ .

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