Gaussian elimination
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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. | ||
| + | |||
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| + | ---- | ||
| + | <i> Return to [[Numerical methods | Numerical Methods]] </i> | ||
Revision as of 06:22, 3 October 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}
{x_1 } \\
{x_2 } \\
. \\
{x_n } \\
\end{matrix}
\right]
=
\left[
\begin{matrix}
{b_1 } \\
{b_2 } \\
. \\
{b_n } \\
\end{matrix}
\right]](/W/images/math/6/7/7/677a1968347218633c4702a422ca4c2a.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}
{x_1 } \\
{x_2 } \\
. \\
{x_n } \\
\end{matrix}
\right]](/W/images/math/3/b/1/3b171f419df579fc32d4184f498753fa.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.
Return to Numerical Methods
