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Gaussian elimination

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m (Gauss elimination moved to Gaussian elimination)
(towards a uniform notation for linear systems : A*Phi = B)
Line 1: Line 1:
== Gauss Elimination ==
== Gauss Elimination ==
-
We consider the system of linear equations '''<math> Ax = b </math>''' or <br>
+
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}
-
   {x_1 }  \\  
+
   {\phi_1 }  \\  
-
   {x_2 }  \\  
+
   {\phi_2 }  \\  
   .  \\
   .  \\
-
   {x_n }  \\
+
   {\phi_n }  \\
\end{matrix}
\end{matrix}
\right]
\right]
Line 52: Line 52:
\left[  
\left[  
\begin{matrix}
\begin{matrix}
-
   {x_1 }  \\  
+
   {\phi_1 }  \\  
-
   {x_2 }  \\  
+
   {\phi_2 }  \\  
   .  \\  
   .  \\  
-
   {x_n }  \\  
+
   {\phi_n }  \\  
\end{matrix}
\end{matrix}
\right]
\right]
Line 86: Line 86:
By using the formula: <br>
By using the formula: <br>
:<math>
:<math>
-
x_i = {1 \over {a_{ii}^' }}\left( {b_i^'  - \sum\limits_{j = i + 1}^n {a_{ij}^' x_j } } \right)
+
\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  A\cdot\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 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:


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

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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