Triangular matrix: Difference between revisions

Revision as of 14:05, 21 January 2005

Revision as of 16:26, 27 January 2005

Line 94:

:<math>\mathbf{U} \mathbf{x} = \mathbf{b}</math>

is very easy to solve. The matrix equation ''Lx'' = ''b'' can be written as a system of linear equations

:<math>

\begin{matrix}

l_{1,1} x_1 & & x_1 & & & & & & & = & b_1 \\

l_{2,1} x_1 & + & l_{2,2} x_2 & x_2 & & & & = & b_2 \\

\vdots & & \vdots & \ddots & & & \vdots \\

l_{m,1} x_1 & + & l_{m,2} x_2 & + \ldots + & l_{m,m} x_m & = & b_m \\

\end{matrix}

</math>

which can be solved by the following recursive relation

:<math> x_1 = \frac{b_1}{l_{1,1}}, </math>

:<math>

:<math> x_2 = \frac{b_2 - l_{2,1} x_1}{l_{2,2}}, </math>

\begin{matrix}

::<math> \vdots </math>

x_1 & = & b_1 \\

x_m:<math> &x_m = & \frac{b_m - \sum_{i=1}^{m-1} l_{m,i}b_ix_i}{l_{m,m}}. </math>

x_2 & = & b_2 - l_{2,1} b_1 \\

& \vdots & \\

x_m & = & b_m - \sum_{i=1}^{m-1} l_{m,i}b_i

\end{matrix}

:</math>

A matrix equation with a normedan upper triangular matrix ''U'' can be solved in an analogous way.

== See also ==