Conjugate gradient numerical (5 iterations)

$A=\begin{bmatrix}3&0&1\\0&-1&3\\1&3&0\end{bmatrix}$, $b=\{1,-12,2\}^T$. $A$ is symmetric but not positive definite (its eigenvalues are $3.61,\,2,\,-3.61$). The CG recurrences can still be applied as long as no denominator $d^TAd$ is zero, but $\alpha$ may then become negative. Take $x_0=0$, so $r_0=b$ and $d_0=r_0=\{1,-12,2\}^T$.

Iteration 1

$$\begin{aligned} Ad_0&=\{5,\,18,\,-35\}^T\\ \alpha_0&=\frac{149}{-281}=-0.53025\\ x_1&=\{-0.5302,\,6.3630,\,-1.0605\}^T\\ r_1&=\{3.6512,\,-2.4555,\,-16.5587\}^T\\ \beta_0&=\frac{293.55}{149}=1.97015,\quad d_1=\{5.6214,\,-26.0973,\,-12.6184\}^T \end{aligned}$$

Iteration 2

$$\begin{aligned} Ad_1&=\{4.2458,\,-11.7579,\,-72.6706\}^T\\ \alpha_1&=\frac{293.55}{1247.7}=0.23527\\ x_2&=\{0.7923,\,0.2230,\,-4.0293\}^T\\ r_2&=\{2.6523,\,0.3108,\,0.5388\}^T\\ \beta_1&=\frac{7.4217}{293.55}=0.025282,\quad d_2=\{2.7944,\,-0.3490,\,0.2197\}^T \end{aligned}$$

Iteration 3

$$\begin{aligned} Ad_2&=\{8.6031,\,1.0082,\,1.7475\}^T\\ \alpha_2&=\frac{7.4217}{24.073}=0.30830\\ x_3&=\{1.6538,\,0.1154,\,-3.9615\}^T\\ r_3&\approx\{0,\,0,\,0\}^T \end{aligned}$$

The residual is zero after three iterations, so iterations 4 and 5 are not needed.

Answer: $x_1=1.6538\ (43/26),\ x_2=0.1154\ (3/26),\ x_3=-3.9615\ (-103/26)$. Check: $3(1.6538)-3.9615=1.0$, $-0.1154-11.8846=-12$, $1.6538+0.3462=2$.