Gaussian elimination

From CFD-Wiki

(Difference between revisions)

Revision as of 07:55, 27 September 2005

Gauss Elimination

We consider the system of linear equations $Ax = b$ or

$\left[ \begin{matrix} {a_{11} } & {a_{12} } & {...} & {a_{1n} } \\ {a_{21} } & {a_{22} } & . & {a_{21} } \\ . & . & . & . \\ {a_{n1} } & {a_{n1} } & . & {a_{nn} } \\ \end{matrix} \right] \left[ \begin{matrix} {x_1 } \\ {x_2 } \\ . \\ {x_n } \\ \end{matrix} \right] = \left[ \begin{matrix} {b_1 } \\ {b_2 } \\ . \\ {b_n } \\ \end{matrix} \right]$

To perform Gaussian elimination starting with the above given system of equations we compose the augmented matrix equation in the form:

$\left[ \begin{matrix} {a_{11} } & {a_{12} } & {...} & {a_{1n} } \\ {a_{21} } & {a_{22} } & . & {a_{21} } \\ . & . & . & . \\ {a_{n1} } & {a_{n1} } & . & {a_{nn} } \\ \end{matrix} \left| \begin{matrix} {b_1 } \\ {b_2 } \\ . \\ {b_n } \\ \end{matrix} \right. \right] \left[ \begin{matrix} {x_1 } \\ {x_2 } \\ . \\ {x_n } \\ \end{matrix} \right]$

After performing elementary raw operations the augmented matrix is put into the upper triangular form:

$\left[ \begin{matrix} {a_{11}^' } & {a_{12}^' } & {...} & {a_{1n}^' } \\ 0 & {a_{22}^' } & . & {a_{2n}^' } \\ . & . & . & . \\ 0 & 0 & . & {a_{nn}^' } \\ \end{matrix} \left| \begin{matrix} {b_1^' } \\ {b_1^' } \\ . \\ {b_1^' } \\ \end{matrix} \right. \right]$

By using the formula:

$x_i = {1 \over {a_{ii}^' }}\left( {b_i^' - \sum\limits_{j = i + 1}^n {a_{ij}^' x_j } } \right)$

Solve the equation of the k^th row for x^k, then substitute back into the equation of the (k-1)^st row to obtain a solution for (k-1)^st raw, and so on till k = 1.

Gaussian elimination

From CFD-Wiki

Revision as of 07:55, 27 September 2005

Gauss Elimination

Views

My wiki

wiki navigation

wiki search

wiki toolbox

@@ Line 28: / Line 28: @@
 \right]
 </math>
+To perform Gaussian elimination starting with the above given system of equations we compose the '''augmented matrix equation''' in the form: <br>
+:<math>
+\left[
+\begin{matrix}
+   {a_{11} } & {a_{12} } & {...} & {a_{1n} }  \\
+   {a_{21} } & {a_{22} } & . & {a_{21} }  \\
+   . & . & . & .  \\
+   {a_{n1} } & {a_{n1} } & . & {a_{nn} }  \\
+\end{matrix}
+\left|
+\begin{matrix}
+   {b_1 }  \\
+   {b_2 }  \\
+   .  \\
+   {b_n }  \\
+\end{matrix}
+\right.
+\right]
+\left[
+\begin{matrix}
+   {x_1 }  \\
+   {x_2 }  \\
+   .  \\
+   {x_n }  \\
+\end{matrix}
+\right]
+</math> <br>
+After performing elementary raw operations the '''augmented matrix''' is put into the upper triangular form: <br>
+:<math>
+\left[
+\begin{matrix}
+   {a_{11}^' } & {a_{12}^' } & {...} & {a_{1n}^' }  \\
+& {a_{22}^' } & . & {a_{2n}^' }  \\
+   . & . & . & .  \\
+& 0 & . & {a_{nn}^' }  \\
+\end{matrix}
+\left|
+\begin{matrix}
+    {b_1^' }  \\
+   {b_1^' }  \\
+   .  \\
+   {b_1^' }  \\
+\end{matrix}
+\right.
+\right]
+</math> <br>
+By using the formula: <br>
+:<math>
+x_i  = {1 \over {a_{ii}^' }}\left( {b_i^'  - \sum\limits_{j = i + 1}^n {a_{ij}^' x_j } } \right)
+</math> <br>
+Solve the equation of the k<sup>th</sup> row for x<sup>k</sup>, then substitute back into the equation of the (k-1)<sup>st</sup> row to obtain a solution for  (k-1)<sup>st</sup> raw, and so on till k = 1.