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Gaussian elimination

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(Algorithm)
(Algorithm)
 
Line 111: Line 111:
Back substitution phase
Back substitution phase
:    for k:= n stepdown until 1 do <br>
:    for k:= n stepdown until 1 do <br>
-
::    for i:= 1 step k-1 <br>
+
::    for i:= 1 step until k-1 do <br>
:::    <math>b_i=b_i-a_{ik}/a_{kk}b_{k}</math> <br>
:::    <math>b_i=b_i-a_{ik}/a_{kk}b_{k}</math> <br>
::    end loop (i) <br>
::    end loop (i) <br>

Latest revision as of 15:08, 22 August 2007

Contents

Description

We consider the system of linear equations  A\phi = 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}
   {\phi_1 }  \\ 
   {\phi_2 }  \\ 
   .  \\
   {\phi_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}
   {\phi_1 }  \\ 
   {\phi_2 }  \\ 
   .  \\ 
   {\phi_n }  \\ 
\end{matrix}
\right]

After performing elementary row 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_2^' }  \\ 
   .  \\
   {b_n^' }  \\ 

\end{matrix}

\right.
\right]

The solution to the original system is found via back substitution. The solution to the last equation is


\phi_n = b_n^'/a_{nn}'.

This result may now be substituted into the second to last equation, allowing us to solve for \phi_{n-1}. Repetition of this substitution process will give us the complete solution vector. The back substitution process may be expressed as


\phi_i  = {1 \over {a_{ii}^' }}\left( {b_i^'  - \sum\limits_{j = i + 1}^n {a_{ij}^' \phi_j } } \right),

where i=n,n-1,\ldots,1.

Algorithm

Forward elimination phase

for k:= 1 step until n-1 do
for i:=k+1 step until n do
m = {{a_{ik} } \over {a_{kk} }}
for j:= 1 step until n do
 a_{ij}=a_{ij}-ma_{kj}
end loop (j)
b_i=b_i-mb_k
end loop (i)
end loop (k)

Back substitution phase

for k:= n stepdown until 1 do
for i:= 1 step until k-1 do
b_i=b_i-a_{ik}/a_{kk}b_{k}
end loop (i)
\phi_{k}=b_{k}/a_{kk}
end loop (k)

Important Considerations

Gaussian elimination is best used for relatively small, relatively full systems of equations. If properly used, it should outperform most iterative methods for these systems. As the system to be solved becomes larger, the overhead associated with the more complicated iterative methods becomes less of an issue, and the iterative methods should outperform Gaussian Elimination. For sparse systems, the use of Gaussian elimination is complicated by the possible introduction of more nonzero entries (fill-in). In any case, it is important to keep in mind that the basic algorithm is vulnerable to accuracy issues, including (but not limited to) the distinct possibility of division by zero at various places in the solution process. In practice, it is best to employ safeguards against such problems (e.g. pivoting).

External link

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