Question

Use Gauss-seidal iteration method to solve the system of equations.
$8x_1-3x_2+2x_3=20,4x_1+11x_2-x_3=33,6x_1-3x_2+12x_3=36$

Answer 1

Lower triangular component L

$\left(\begin{array}{ccc}8& 0& 0\\ 4& 11& 0\\ 6& -3& 12\end{array}\right)$

Upper triangular component L

$\left(\begin{array}{ccc}0& -3& 2\\ 0& 0& -1\\ 0& 0& 0\end{array}\right)$

Inverse of $L^{-1}$

$\left(\begin{array}{ccc}0.125& 0& 0\\ -0.0455& 0.0909& 0\\ -0.0739& 0.0227& 0.0833\end{array}\right)$

Calculation of T

$\left(\begin{array}{ccc}0.125& 0& 0\\ -0.0455& 0.0909& 0\\ -0.0739& 0.0227& 0.0833\end{array}\right)\times \left(\begin{array}{ccc}0& -3& 2\\ 0& 0& -1\\ 0& 0& 0\end{array}\right)=\left(\begin{array}{ccc}0& 0.375& -0.25\\ 0& -0.1364& 0.1818\\ 0& -0.2216& 0.1705\end{array}\right)$

Calculation of C

$\left(\begin{array}{ccc}0.125& 0& 0\\ -0.0455& 0.0909& 0\\ -0.0739& 0.0227& 0.0833\end{array}\right)\times \left(\begin{array}{c}20\\ 33\\ 36\end{array}\right)=\left(\begin{array}{c}2.5\\ 2.0909\\ 2.2727\end{array}\right)$

Gauss Seidel Algorithm

Suppose

$x^{\left(0\right)}=\left(\begin{array}{c}2.5\\ 2.0909\\ 2.2727\end{array}\right)$

$x^{\left(1\right)}=\left(\begin{array}{ccc}0& 0.375& -0.25\\ 0& -0.1364& 0.1818\\ 0& -0.2216& 0.1705\end{array}\right)\times \left(\begin{array}{c}2.5\\ 2.0909\\ 2.2727\end{array}\right)+\left(\begin{array}{c}2.5\\ 2.0909\\ 2.2727\end{array}\right)=\left(\begin{array}{c}2.7159\\ 2.2189\\ 2.1969\end{array}\right)$

$x^{\left(2\right)}=\left(\begin{array}{ccc}0& 0.375& -0.25\\ 0& -0.1364& 0.1818\\ 0& -0.2216& 0.1705\end{array}\right)\times \left(\begin{array}{c}2.7159\\ 2.2189\\ 2.1969\end{array}\right)+\left(\begin{array}{c}2.5\\ 2.0909\\ 2.2727\end{array}\right)=\left(\begin{array}{c}2.7829\\ 2.1876\\ 2.1556\end{array}\right)$

$x^{\left(3\right)}=\left(\begin{array}{ccc}0& 0.375& -0.25\\ 0& -0.1364& 0.1818\\ 0& -0.2216& 0.1705\end{array}\right)\times \left(\begin{array}{c}2.7829\\ 2.1876\\ 2.1556\end{array}\right)+\left(\begin{array}{c}2.5\\ 2.0909\\ 2.2727\end{array}\right)=\left(\begin{array}{c}2.7815\\ 2.1844\\ 2.1555\end{array}\right)$

$x^{\left(4\right)}=\left(\begin{array}{ccc}0& 0.375& -0.25\\ 0& -0.1364& 0.1818\\ 0& -0.2216& 0.1705\end{array}\right)\times \left(\begin{array}{c}2.7815\\ 2.1844\\ 2.1555\end{array}\right)+\left(\begin{array}{c}2.5\\ 2.0909\\ 2.2727\end{array}\right)=\left(\begin{array}{c}2.7803\\ 2.1848\\ 2.1561\end{array}\right)$

$x^{\left(5\right)}=\left(\begin{array}{ccc}0& 0.375& -0.25\\ 0& -0.1364& 0.1818\\ 0& -0.2216& 0.1705\end{array}\right)\times \left(\begin{array}{c}2.7803\\ 2.1848\\ 2.1561\end{array}\right)+\left(\begin{array}{c}2.5\\ 2.0909\\ 2.2727\end{array}\right)=\left(\begin{array}{c}2.7803\\ 2.1849\\ 2.1562\end{array}\right)$

$x^{\left(6\right)}=\left(\begin{array}{ccc}0& 0.375& -0.25\\ 0& -0.1364& 0.1818\\ 0& -0.2216& 0.1705\end{array}\right)\times \left(\begin{array}{c}2.7803\\ 2.1849\\ 2.1562\end{array}\right)+\left(\begin{array}{c}2.5\\ 2.0909\\ 2.2727\end{array}\right)=\left(\begin{array}{c}2.7803\\ 2.1849\\ 2.1562\end{array}\right)$

Result

$x=A^{-1}b=\left(\begin{array}{c}2.7803\\ 2.1849\\ 2.1562\end{array}\right)$

$x_1=2.7803,x_2=2.1849,x_3=2.1562$