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Past exam of the year 1 of the computer science course of the University of Oxford

2017

Digital systems, linear algebra, introduction to formal proof

The exact solution obeys $(D-L)x=Ux+b$. Subtract this from the iteration to get $(D-L)e_n=Ue_{n-1}$. Hence the <stationary iteration matrix> is
$$
\boxed{G=(D-L)^{-1}U
=\begin{pmatrix}1/4&0\\-1/8&1/2\end{pmatrix}
\begin{pmatrix}0&-3\\0&0\end{pmatrix}
=\begin{pmatrix}0&-3/4\\0&3/8\end{pmatrix},\qquad e_n=Ge_{n-1}.}
$$
This is exactly the matrix $B$ from part (a); the sign follows the given $A=D-L-U$ convention.

Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Digital systems, linear algebra, introduction to formal proof / 6 / b / iii / Solution

Past exam of the year 1 of the computer science course of the University of Oxford / 2017 / Digital systems, linear algebra, introduction to formal proof / 6 / b / iii / Solution

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