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 $x_2=(1/4,3/4,-3/4{)}^T$.