Gaussian elimination
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== Gauss Elimination == | == Gauss Elimination == | ||
| - | We consider the system of linear equations '''<math> | + | We consider the system of linear equations '''<math> A\cdot\phi = B </math>''' or <br> |
:<math> | :<math> | ||
\left[ | \left[ | ||
| Line 12: | Line 12: | ||
\left[ | \left[ | ||
\begin{matrix} | \begin{matrix} | ||
| - | { | + | {\phi_1 } \\ |
| - | { | + | {\phi_2 } \\ |
. \\ | . \\ | ||
| - | { | + | {\phi_n } \\ |
\end{matrix} | \end{matrix} | ||
\right] | \right] | ||
| Line 52: | Line 52: | ||
\left[ | \left[ | ||
\begin{matrix} | \begin{matrix} | ||
| - | { | + | {\phi_1 } \\ |
| - | { | + | {\phi_2 } \\ |
. \\ | . \\ | ||
| - | { | + | {\phi_n } \\ |
\end{matrix} | \end{matrix} | ||
\right] | \right] | ||
| Line 86: | Line 86: | ||
By using the formula: <br> | By using the formula: <br> | ||
:<math> | :<math> | ||
| - | + | \phi_i = {1 \over {a_{ii}^' }}\left( {b_i^' - \sum\limits_{j = i + 1}^n {a_{ij}^' \phi_j } } \right) | |
</math> <br> | </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. | 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. | ||
Revision as of 20:37, 15 December 2005
Gauss Elimination
We consider the system of linear equations 
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}
{\phi_1 } \\
{\phi_2 } \\
. \\
{\phi_n } \\
\end{matrix}
\right]
=
\left[
\begin{matrix}
{b_1 } \\
{b_2 } \\
. \\
{b_n } \\
\end{matrix}
\right]](/W/images/math/a/4/d/a4dada28dd7a85c53604dc331829fc4e.png)
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}
{\phi_1 } \\
{\phi_2 } \\
. \\
{\phi_n } \\
\end{matrix}
\right]](/W/images/math/4/7/e/47e9bee10079eb1421bd67e9bd451eaf.png)
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]](/W/images/math/c/f/9/cf9d0e176e3916d1cb1a925ca3b34e00.png)
By using the formula:
Solve the equation of the kth row for xk, 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.
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