Question A.12: Convert the matrix of Example A.10 into LDU form |1 2 1 0 0 ......
Convert the matrix of Example A.10 into LDU form
\left|\begin{matrix}1 &2 &1 &0 \\ 0& 3& 3& 1 \\ 2& 0& 2& 0 \\ 1 &0 &0 &2 \end{matrix} \right| = l¹ × l² × l³ × D × u³ × u² × u¹
The lower matrices are
l¹ × l² × l³ = \left|\begin{matrix}1 &0&0 &0 \\ 0& 1& 0&0 \\ 2& 0& 1& 0 \\ 1 &0 &0 &1 \end{matrix} \right| \left|\begin{matrix}1 &0&0 &0 \\ 0& 1& 0&0 \\ 0& -4/3& 1& 0 \\ 1 &-2/3 &0 &1 \end{matrix} \right| \left|\begin{matrix}1 &0&0 &0 \\ 0& 1& 0&0 \\ 0& 0& 1& 0 \\ 0&0 &1/4 &0 \end{matrix} \right|
The upper matrices are
u³ × u² × u¹ = \left|\begin{matrix}1 &0&0 &0 \\ 0& 1& 0&1/3 \\ 0& 0& 1& 1/3 \\ 0 &0 &0&1 \end{matrix} \right| \left|\begin{matrix}1 &0&0 &0 \\ 0& 1& 1&0 \\ 0& 0& 1& 0 \\ 0&0 &0 &1 \end{matrix} \right| \left|\begin{matrix}1 &2&1 &0 \\ 0& 1& 0&0 \\ 0& 0& 1& 0 \\ 0&0 &0 &1 \end{matrix} \right|
The matrix D is
D = \left|\begin{matrix}1 &0&0 &0 \\ 0& 3& 0&0 \\ 0& 0& 4& 0 \\ 0&0 &0 &7/3 \end{matrix} \right|Thus, the LDU form of the original matrix is
If the coefficient matrix is symmetrical (for a linear bilateral network), then
[L]=[U]^{t} (A.109)
Because
l_{ip}(new) = a_{ip}/a_{pp}
u_{pi} = a_{pi}/a_{pp} (a_{ip} = a_{pi}) (A.110)
The LU and LDU forms are extensively used in power systems.