 Solution

From the zero initial vector, the linear-system residual is $r_0=(1,1{)}^T$. Then $r_0^T{r}_0=2$, $A{r}_0=(5,13{)}^T$, and $r_0^{T}A{r}_0=18$. The step is ${\alpha }_0=1/9$, givingAs 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.