Conjugate gradient numerical (2,-1 system)

Matrix form: $A=\begin{bmatrix}2&-1&0\\-1&2&-1\\0&-1&2\end{bmatrix}$, $b=\{4,0,0\}^T$. $A$ is symmetric and positive definite. Start with $x_0=\{0,0,0\}^T$.

$r_0=b-Ax_0=\{4,0,0\}^T$, $d_0=r_0$.

Iteration 1

$$\begin{aligned} Ad_0&=\{8,-4,0\}^T\\ \alpha_0&=\frac{r_0^Tr_0}{d_0^TAd_0}=\frac{16}{32}=0.5\\ x_1&=\{2,0,0\}^T\\ r_1&=r_0-\alpha_0Ad_0=\{0,2,0\}^T\\ \beta_0&=\frac{4}{16}=0.25,\quad d_1=r_1+\beta_0d_0=\{1,2,0\}^T \end{aligned}$$

Iteration 2

$$\begin{aligned} Ad_1&=\{0,3,-2\}^T\\ \alpha_1&=\frac{4}{6}=0.6667\\ x_2&=\{2.6667,\,1.3333,\,0\}^T\\ r_2&=\{0,0,1.3333\}^T\\ \beta_1&=\frac{1.7778}{4}=0.4444,\quad d_2=\{0.4444,\,0.8889,\,1.3333\}^T \end{aligned}$$

Iteration 3

$$\begin{aligned} Ad_2&=\{0,0,1.7778\}^T\\ \alpha_2&=\frac{1.7778}{2.3704}=0.75\\ x_3&=\{3,\,2,\,1\}^T\\ r_3&=\{0,0,0\}^T \end{aligned}$$

The residual is zero after $n=3$ iterations, as expected for CG.

Answer: $x_1=3,\ x_2=2,\ x_3=1$. (Check: $2(3)-2=4$, $-3+4-1=0$, $-2+2=0$.)