Find three sets of iterative solutions by the Gauss–Jacobi method: \(8x+y+2z=16,\ x+6y-z=10,\ 2x-y+9z=27\).

Answer: \((2,\,1.6667,\,3)\), \((1.0417,\,1.8333,\,2.7407)\), \((1.0856,\,1.9498,\,2.9722)\); the iterates approach the exact solution \((1,2,3)\).

Diagonally dominant: \(8>1+2,\ 6>1+1,\ 9>2+1\).

\[x=\frac{16-y-2z}{8},\quad y=\frac{10-x+z}{6},\quad z=\frac{27-2x+y}{9}\]

In Jacobi's method every new value uses only the values of the previous iteration. Start with \(x=y=z=0\).

Iteration 1: \(x=\frac{16}{8}=2\), \(y=\frac{10}{6}=1.6667\), \(z=\frac{27}{9}=3\).

Iteration 2: \(x=\frac{16-1.6667-6}{8}=1.0417\), \(y=\frac{10-2+3}{6}=1.8333\), \(z=\frac{27-4+1.6667}{9}=2.7407\).

Iteration 3: \(x=\frac{16-1.8333-5.4815}{8}=1.0856\), \(y=\frac{10-1.0417+2.7407}{6}=1.9498\), \(z=\frac{27-2.0833+1.8333}{9}=2.9722\).

Check: \((1,2,3)\) satisfies \(8+2+6=16\), \(1+12-3=10\), \(2-2+27=27\). ✓