Then A can be decomposed into a diagonal component D, and strictly lower and upper triangular components L and U: $$ \mathbf{A} =\mathbf{D} + \mathbf{L} + \mathbf{U}, $$ where $$ D = \begin{bmatrix} a_{11} & 0 & \cdots & 0 \\ 0 & a_{22} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & a_{nn} \end{bmatrix}, \quad L = \begin{bmatrix} 0 & 0 & \cdots & 0 \\ a_{21} & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\a_{n1} & a_{n2} & \cdots & 0 \end{bmatrix}, \quad U = \begin{bmatrix} 0 & a_{12} & \cdots & a_{1n} \\ 0 & 0 & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\0 & 0 & \cdots & 0 \end{bmatrix}. $$ The system of linear equations may be rewritten as: $$ (D+\omega L) \mathbf{x} = \omega \mathbf{b} - [\omega U + (\omega-1) D ] \mathbf{x} $$ for a constant \( \omega > 1 \).