Solution
= Solution
From the zero initial <vector>, the <linear-system residual> is $r_0=(1,1)^T$. Then $r_0^Tr_0=2$, $Ar_0=(5,13)^T$, and $r_0^TAr_0=18$. The step is $\alpha_0=1/9$, giving
$$
\boxed{x_1=\begin{pmatrix}1/9\\1/9\end{pmatrix}.}
$$
As a check, the new residual is $(4/9,-4/9)^T$, orthogonal to the preceding residual as expected after this exact line search. It is nonzero, so one iteration has not already solved the system.