Solve the following system by the Gauss–Seidel method (three iterations): \(10x+2y+z=9,\ 2x+20y-2z=-44,\ -2x+3y+10z=22\).

Answer: After three iterations \(x\approx0.9987,\ y\approx-1.9989,\ z\approx2.9994\) (exact solution \(x=1,\ y=-2,\ z=3\)).

The system is diagonally dominant (\(10>2+1,\ 20>2+2,\ 10>2+3\)), so Gauss–Seidel converges.

\[x=\frac{9-2y-z}{10},\quad y=\frac{-44-2x+2z}{20},\quad z=\frac{22+2x-3y}{10}\]

Start with \(y=z=0\) and always use the latest values.

Iteration 1: \(x=\frac{9}{10}=0.9\); \(y=\frac{-44-1.8+0}{20}=-2.29\); \(z=\frac{22+1.8+6.87}{10}=3.067\).

Iteration 2: \(x=\frac{9+4.58-3.067}{10}=1.0513\); \(y=\frac{-44-2.1026+6.134}{20}=-1.9984\); \(z=\frac{22+2.1026+5.9953}{10}=3.0098\).

Iteration 3: \(x=\frac{9+3.9969-3.0098}{10}=0.9987\); \(y=\frac{-44-1.9974+6.0196}{20}=-1.9989\); \(z=\frac{22+1.9974+5.9967}{10}=2.9994\).

The values are approaching \((1,-2,3)\), which satisfies all three equations.