Eliminación de Gauss

Eliminación hacia atrás

Transformar el sistema de ecuaciones en un sistema triangular inferior

\begin{equation*} \begin{bmatrix} 3 & -0.1 & -0.2 \\ 0.1 & 7 & -0.3 \\ 0.3 & -0.2 & 10 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 7.85 \\ -19.3 \\ 71.4 \end{bmatrix} \end{equation*}

Tercera fila pivote

\begin{equation*} \begin{bmatrix} 3 - \frac{-0.2}{10} (0.3) & -0.1 - \frac{-0.2}{10} (-0.2) & -0.2 - \frac{-0.2}{10} (10) \\ 0.1 - \frac{-0.3}{10} (0.3) & 7 - \frac{-0.3}{10} (-0.2) & -0.3 - \frac{-0.3}{10} (10) \\ 0.3 & -0.2 & 10 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 7.85 - \frac{-0.2}{10} (71.4) \\ -19.3 - \frac{-0.3}{10} (71.4) \\ 71.4 \end{bmatrix} \end{equation*}

Simplificando

\begin{equation*} \begin{bmatrix} 3.006 & -0.104 & 0 \\ 0.109 & 6.994 & 0 \\ 0.3 & -0.2 & 10 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 9.278 \\ -17.158 \\ 71.4 \end{bmatrix} \end{equation*}

Segunda fila pivote

\begin{equation*} \begin{bmatrix} 3.006 - \frac{-0.104}{6.994} (0.109) & -0.104 - \frac{-0.104}{6.994} (6.994) & 0 \\ 0.109 & 6.994 & 0 \\ 0.3 & -0.2 & 10 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 9.278 - \frac{-0.104}{6.994} (-17.158) \\ -17.158 \\ 71.4 \end{bmatrix} \end{equation*}

Simplificando

\begin{equation*} \begin{bmatrix} 3.007621 & 0 & 0 \\ 0.109 & 6.994 & 0 \\ 0.3 & -0.2 & 10 \end{bmatrix} \begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} 9.022862 \\ -17.158 \\ 71.4 \end{bmatrix} \end{equation*}

Algoritmo de cálculo

\begin{align*} a_{11} x_{1} + a_{12} x_{2} + a_{13} x_{3} + a_{14} x_{4} &= b_{1} \\ a_{21} x_{1} + a_{22} x_{2} + a_{23} x_{3} + a_{24} x_{4} &= b_{2} \\ a_{31} x_{1} + a_{32} x_{2} + a_{33} x_{3} + a_{34} x_{4} &= b_{3} \\ a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} &= b_{4} \end{align*}

Cuarta fila pivote

\begin{align*} a_{11} x_{1} + a_{12} x_{2} + a_{13} x_{3} + a_{14} x_{4} - \frac{a_{14}}{a_{44}} ( a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} ) &= b_{1} - \frac{a_{14}}{a_{44}} b_{4} \\ a_{21} x_{1} + a_{22} x_{2} + a_{23} x_{3} + a_{24} x_{4} - \frac{a_{24}}{a_{44}} ( a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} ) &= b_{2} - \frac{a_{24}}{a_{44}} b_{4} \\ a_{21} x_{1} + a_{22} x_{2} + a_{23} x_{3} + a_{34} x_{4} - \frac{a_{34}}{a_{44}} ( a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} ) &= b_{3} - \frac{a_{34}}{a_{44}} b_{4} \\ a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} &= b_{4} \end{align*}

Expandiendo y agrupando términos

\begin{align*} \bigg( a_{11} - \frac{a_{13}}{a_{33}} a_{41} \bigg) x_{1} + \bigg( a_{12} - \frac{a_{13}}{a_{33}} a_{42} \bigg) x_{2} + \bigg( a_{13} - \frac{a_{13}}{a_{33}} a_{43} \bigg) x_{3} + \bigg( a_{14} - \frac{a_{14}}{a_{33}} a_{44} \bigg) x_{4} &= b_{1} - \frac{a_{14}}{a_{44}} b_{4} \\ \bigg( a_{21} - \frac{a_{23}}{a_{33}} a_{41} \bigg) x_{1} + \bigg( a_{22} - \frac{a_{23}}{a_{33}} a_{42} \bigg) x_{2} + \bigg( a_{23} - \frac{a_{23}}{a_{33}} a_{43} \bigg) x_{3} + \bigg( a_{24} - \frac{a_{24}}{a_{33}} a_{44} \bigg) x_{4} &= b_{2} - \frac{a_{24}}{a_{44}} b_{4} \\ \bigg( a_{31} - \frac{a_{23}}{a_{33}} a_{41} \bigg) x_{1} + \bigg( a_{32} - \frac{a_{23}}{a_{33}} a_{42} \bigg) x_{2} + \bigg( a_{33} - \frac{a_{23}}{a_{33}} a_{43} \bigg) x_{3} + \bigg( a_{34} - \frac{a_{34}}{a_{33}} a_{44} \bigg) x_{4} &= b_{3} - \frac{a_{34}}{a_{44}} b_{4} \\ a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} &= b_{4} \end{align*}

Usando un cambio de variable

\begin{alignedat}{5} a'_{11} x_{1} & {}+{} a'_{12} x_{2} & {}+{} a'_{13} x_{3} & & {}={} b'_{1} \\ a'_{21} x_{1} & {}+{} a'_{22} x_{2} & {}+{} a'_{23} x_{3} & & {}={} b'_{2} \\ a'_{31} x_{1} & {}+{} a'_{32} x_{2} & {}+{} a'_{33} x_{3} & & {}={} b'_{3} \\ a_{41} x_{1} & {}+{} a_{42} x_{2} & {}+{} a_{43} x_{3} & {}+{} a_{44} x_{4} & {}={} b_{4} \end{alignedat}

Tercera fila pivote

\begin{align*} a'_{11} x_{1} + a'_{12} x_{2} + a'_{13} x_{3} - \frac{a'_{13}}{a'_{33}} ( a'_{31} x_{1} + a'_{32} x_{2} + a'_{33} x_{3} ) &= b'_{1} - \frac{a'_{13}}{a'_{33}} b'_{3} \\ a'_{21} x_{1} + a'_{22} x_{2} + a'_{23} x_{3} - \frac{a'_{23}}{a'_{33}} ( a'_{31} x_{1} + a'_{32} x_{2} + a'_{33} x_{3} ) &= b'_{2} - \frac{a'_{23}}{a'_{33}} b'_{3} \\ a'_{31} x_{1} + a'_{32} x_{2} + a'_{33} x_{3} &= b'_{3} \\ a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} &= b_{4} \end{align*}

Expandiendo y agrupando términos

\begin{align*} \bigg( a'_{11} - \frac{a'_{13}}{a'_{33}} a'_{31} \bigg) x_{1} + \bigg( a'_{12} - \frac{a'_{13}}{a'_{33}} a'_{32} \bigg) x_{2} + \bigg( a'_{13} - \frac{a'_{13}}{a'_{33}} a'_{33} \bigg) x_{3} &= b'_{1} - \frac{a'_{13}}{a'_{33}} b'_{3} \\ \bigg( a'_{21} - \frac{a'_{23}}{a'_{33}} a'_{31} \bigg) x_{1} + \bigg( a'_{22} - \frac{a'_{23}}{a'_{33}} a'_{32} \bigg) x_{2} + \bigg( a'_{23} - \frac{a'_{23}}{a'_{33}} a'_{33} \bigg) x_{3} &= b'_{2} - \frac{a'_{23}}{a'_{33}} b'_{3} \\ a'_{31} x_{1} + a'_{32} x_{2} + a'_{33} x_{3} &= b'_{3} \\ a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} &= b_{4} \end{align*}

Usando un cambio de variable

\begin{alignedat}{5} a''_{11} x_{1} & {}+{} a''_{12} x_{2} & & & {}={} b''_{1} \\ a''_{21} x_{1} & {}+{} a''_{22} x_{2} & & & {}={} b''_{2} \\ a'_{31} x_{1} & {}+{} a'_{32} x_{2} & {}+{} a'_{33} x_{3} & & {}={} b'_{3} \\ a_{41} x_{1} & {}+{} a_{42} x_{2} & {}+{} a_{43} x_{3} & {}+{} a_{44} x_{4} & {}={} b_{4} \end{alignedat}

Segunda fila pivote

\begin{align*} a''_{11} x_{1} + a''_{12} x_{2} - \frac{a''_{12}}{a''_{22}} ( a''_{21} x_{1} + a''_{22} x_{2} ) &= b''_{1} - \frac{a''_{12}}{a''_{22}} b''_{2} \\ a''_{21} x_{1} + a''_{22} x_{2} &= b''_{2} \\ a'_{31} x_{1} + a'_{32} x_{2} + a'_{33} x_{3} &= b'_{3} \\ a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} &= b_{4} \end{align*}

Expandiendo y agrupando términos

\begin{alignedat}{5} \bigg( a''_{11} - \frac{a''_{12}}{a''_{22}} a''_{21} \bigg) x_{1} + \bigg( a''_{12} - \frac{a''_{12}}{a''_{22}} a''_{22} \bigg) x_{2} &= b''_{1} - \frac{a''_{12}}{a''_{22}} b''_{2} \\ a''_{21} x_{1} + a''_{22} x_{2} &= b''_{2} \\ a'_{31} x_{1} + a'_{32} x_{2} + a'_{33} x_{3} &= b'_{3} \\ a_{41} x_{1} + a_{42} x_{2} + a_{43} x_{3} + a_{44} x_{4} &= b_{4} \end{alignedat}

Usando un cambio de variable

\begin{alignedat}{5} a'''_{11} x_{1} & & & & {}={} b'''_{1} \\ a''_{21} x_{1} & {}+{} a''_{22} x_{2} & & & {}={} b''_{2} \\ a'_{31} x_{1} & {}+{} a'_{32} x_{2} & {}+{} a'_{33} x_{3} & & {}={} b'_{3} \\ a_{41} x_{1} & {}+{} a_{42} x_{2} & {}+{} a_{43} x_{3} & {}+{} a_{44} x_{4} & {}={} b_{4} \end{alignedat}

Lo anterior puede escribirse como

\begin{equation*} \begin{array}{llll:l} a_{11} = a_{11} - \frac{a_{14}}{a_{44}} a_{41} & a_{12} = a_{12} - \frac{a_{14}}{a_{44}} a_{42} & a_{13} = a_{13} - \frac{a_{14}}{a_{44}} a_{43} & a_{14} = a_{14} - \frac{a_{14}}{a_{44}} a_{44} & b_{1} = b_{1} - \frac{a_{14}}{a_{44}} b_{4} \\ a_{21} = a_{21} - \frac{a_{24}}{a_{44}} a_{41} & a_{22} = a_{22} - \frac{a_{24}}{a_{44}} a_{42} & a_{23} = a_{23} - \frac{a_{24}}{a_{44}} a_{43} & a_{24} = a_{24} - \frac{a_{24}}{a_{44}} a_{44} & b_{2} = b_{2} - \frac{a_{24}}{a_{44}} b_{4} \\ a_{31} = a_{31} - \frac{a_{34}}{a_{44}} a_{41} & a_{32} = a_{32} - \frac{a_{34}}{a_{44}} a_{42} & a_{33} = a_{33} - \frac{a_{34}}{a_{44}} a_{43} & a_{34} = a_{34} - \frac{a_{34}}{a_{44}} a_{44} & b_{3} = b_{3} - \frac{a_{34}}{a_{44}} b_{4} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \\ \hline a_{11} = a_{11} - \frac{a_{13}}{a_{33}} a_{31} & a_{12} = a_{12} - \frac{a_{13}}{a_{33}} a_{32} & a_{13} = a_{13} - \frac{a_{13}}{a_{33}} a_{33} & a_{14} = a_{14} - \frac{a_{13}}{a_{33}} a_{34} & b_{1} = b_{1} - \frac{a_{13}}{a_{33}} b_{3} \\ a_{21} = a_{21} - \frac{a_{23}}{a_{33}} a_{31} & a_{22} = a_{22} - \frac{a_{23}}{a_{33}} a_{32} & a_{23} = a_{23} - \frac{a_{23}}{a_{33}} a_{33} & a_{24} = a_{24} - \frac{a_{23}}{a_{33}} a_{34} & b_{2} = b_{2} - \frac{a_{23}}{a_{33}} b_{3} \\ a_{31} = a_{31} & a_{32} = a_{32} & a_{33} = a_{33} & a_{34} = a_{34} & b_{3} = b_{3} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \\ \hline a_{11} = a_{11} - \frac{a_{12}}{a_{22}} a_{21} & a_{12} = a_{12} - \frac{a_{12}}{a_{22}} a_{22} & a_{13} = a_{13} - \frac{a_{12}}{a_{22}} a_{23} & a_{14} = a_{14} - \frac{a_{12}}{a_{22}} a_{24} & b_{1} = b_{1} - \frac{a_{12}}{a_{22}} b_{2} \\ a_{21} = a_{21} & a_{22} = a_{22} & a_{23} = a_{23} & a_{24} = a_{24} & b_{2} = b_{2} \\ a_{31} = a_{31} & a_{32} = a_{32} & a_{33} = a_{33} & a_{34} = a_{34} & b_{3} = b_{3} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \end{array} \end{equation*}

Patrón de cálculo

Primer patrón

\begin{equation*} \begin{array}{llll:l} a_{1\color{fuchsia}{1}} = a_{1\color{fuchsia}{1}} - \frac{a_{14}}{a_{44}} a_{4\color{fuchsia}{1}} & a_{1\color{red}{2}} = a_{1\color{red}{2}} - \frac{a_{14}}{a_{44}} a_{4\color{red}{2}} & a_{1\color{green}{3}} = a_{1\color{green}{3}} - \frac{a_{14}}{a_{44}} a_{4\color{green}{3}} & a_{1\color{blue}{4}} = a_{1\color{blue}{4}} - \frac{a_{14}}{a_{44}} a_{4\color{blue}{4}} & b_{1} = b_{1} - \frac{a_{14}}{a_{44}} b_{4} \\ a_{2\color{fuchsia}{1}} = a_{2\color{fuchsia}{1}} - \frac{a_{24}}{a_{44}} a_{4\color{fuchsia}{1}} & a_{2\color{red}{2}} = a_{2\color{red}{2}} - \frac{a_{24}}{a_{44}} a_{4\color{red}{2}} & a_{2\color{green}{3}} = a_{2\color{green}{3}} - \frac{a_{24}}{a_{44}} a_{4\color{green}{3}} & a_{2\color{blue}{4}} = a_{2\color{blue}{4}} - \frac{a_{24}}{a_{44}} a_{4\color{blue}{4}} & b_{2} = b_{2} - \frac{a_{24}}{a_{44}} b_{4} \\ a_{3\color{fuchsia}{1}} = a_{3\color{fuchsia}{1}} - \frac{a_{34}}{a_{44}} a_{4\color{fuchsia}{1}} & a_{3\color{red}{2}} = a_{3\color{red}{2}} - \frac{a_{34}}{a_{44}} a_{4\color{red}{2}} & a_{3\color{green}{3}} = a_{3\color{green}{3}} - \frac{a_{34}}{a_{44}} a_{4\color{green}{3}} & a_{3\color{blue}{4}} = a_{3\color{blue}{4}} - \frac{a_{34}}{a_{44}} a_{4\color{blue}{4}} & b_{3} = b_{3} - \frac{a_{34}}{a_{44}} b_{4} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \\ \hline a_{1\color{fuchsia}{1}} = a_{1\color{fuchsia}{1}} - \frac{a_{13}}{a_{33}} a_{3\color{fuchsia}{1}} & a_{1\color{red}{2}} = a_{1\color{red}{2}} - \frac{a_{13}}{a_{33}} a_{3\color{red}{2}} & a_{1\color{green}{3}} = a_{1\color{green}{3}} - \frac{a_{13}}{a_{33}} a_{3\color{green}{3}} & a_{1\color{blue}{4}} = a_{1\color{blue}{4}} - \frac{a_{13}}{a_{33}} a_{3\color{blue}{4}} & b_{1} = b_{1} - \frac{a_{13}}{a_{33}} b_{3} \\ a_{2\color{fuchsia}{1}} = a_{2\color{fuchsia}{1}} - \frac{a_{23}}{a_{33}} a_{3\color{fuchsia}{1}} & a_{2\color{red}{2}} = a_{2\color{red}{2}} - \frac{a_{23}}{a_{33}} a_{3\color{red}{2}} & a_{2\color{green}{3}} = a_{2\color{green}{3}} - \frac{a_{23}}{a_{33}} a_{3\color{green}{3}} & a_{2\color{blue}{4}} = a_{2\color{blue}{4}} - \frac{a_{23}}{a_{33}} a_{3\color{blue}{4}} & b_{2} = b_{2} - \frac{a_{23}}{a_{33}} b_{3} \\ a_{31} = a_{31} & a_{32} = a_{32} & a_{33} = a_{33} & a_{34} = a_{34} & b_{3} = b_{3} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \\ \hline a_{1\color{fuchsia}{1}} = a_{1\color{fuchsia}{1}} - \frac{a_{12}}{a_{22}} a_{2\color{fuchsia}{1}} & a_{1\color{red}{2}} = a_{1\color{red}{2}} - \frac{a_{12}}{a_{22}} a_{2\color{red}{2}} & a_{1\color{green}{3}} = a_{1\color{green}{3}} - \frac{a_{12}}{a_{22}} a_{2\color{green}{3}} & a_{1\color{blue}{4}} = a_{1\color{blue}{4}} - \frac{a_{12}}{a_{22}} a_{2\color{blue}{4}} & b_{1} = b_{1} - \frac{a_{12}}{a_{22}} b_{2} \\ a_{21} = a_{21} & a_{22} = a_{22} & a_{23} = a_{23} & a_{24} = a_{24} & b_{2} = b_{2} \\ a_{31} = a_{31} & a_{32} = a_{32} & a_{33} = a_{33} & a_{34} = a_{34} & b_{3} = b_{3} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \end{array} \end{equation*}

lo anterior puede escribirse como

\begin{equation*} a_{?j} = a_{?j} - \frac{a_{??}}{a_{??}} a_{?j} \end{equation*}

para $j = 4, 3, 2, 1 = n , \dots, 1$

Segundo patrón

\begin{equation*} \begin{array}{llll:l} a_{\color{red}{1}1} = a_{\color{red}{1}1} - \frac{a_{\color{red}{1}4}}{a_{44}} a_{41} & a_{\color{red}{1}2} = a_{\color{red}{1}2} - \frac{a_{\color{red}{1}4}}{a_{44}} a_{42} & a_{\color{red}{1}3} = a_{\color{red}{1}3} - \frac{a_{\color{red}{1}4}}{a_{44}} a_{43} & a_{\color{red}{1}4} = a_{\color{red}{1}4} - \frac{a_{\color{red}{1}4}}{a_{44}} a_{44} & b_{\color{red}{1}} = b_{\color{red}{1}} - \frac{a_{\color{red}{1}4}}{a_{44}} b_{4} \\ a_{\color{green}{2}1} = a_{\color{green}{2}1} - \frac{a_{\color{green}{2}4}}{a_{44}} a_{41} & a_{\color{green}{2}2} = a_{\color{green}{2}2} - \frac{a_{\color{green}{2}4}}{a_{44}} a_{42} & a_{\color{green}{2}3} = a_{\color{green}{2}3} - \frac{a_{\color{green}{2}4}}{a_{44}} a_{43} & a_{\color{green}{2}4} = a_{\color{green}{2}4} - \frac{a_{\color{green}{2}4}}{a_{44}} a_{44} & b_{\color{green}{2}} = b_{\color{green}{2}} - \frac{a_{\color{green}{2}4}}{a_{44}} b_{4} \\ a_{\color{blue}{3}1} = a_{\color{blue}{3}1} - \frac{a_{\color{blue}{3}4}}{a_{44}} a_{41} & a_{\color{blue}{3}2} = a_{\color{blue}{3}2} - \frac{a_{\color{blue}{3}4}}{a_{44}} a_{42} & a_{\color{blue}{3}3} = a_{\color{blue}{3}3} - \frac{a_{\color{blue}{3}4}}{a_{44}} a_{43} & a_{\color{blue}{3}4} = a_{\color{blue}{3}4} - \frac{a_{\color{blue}{3}4}}{a_{44}} a_{44} & b_{\color{blue}{3}} = b_{\color{blue}{3}} - \frac{a_{\color{blue}{3}4}}{a_{44}} b_{4} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \\ \hline a_{\color{red}{1}1} = a_{\color{red}{1}1} - \frac{a_{\color{red}{1}3}}{a_{33}} a_{31} & a_{\color{red}{1}2} = a_{\color{red}{1}2} - \frac{a_{\color{red}{1}3}}{a_{33}} a_{32} & a_{\color{red}{1}3} = a_{\color{red}{1}3} - \frac{a_{\color{red}{1}3}}{a_{33}} a_{33} & a_{\color{red}{1}4} = a_{\color{red}{1}4} - \frac{a_{\color{red}{1}3}}{a_{33}} a_{34} & b_{\color{red}{1}} = b_{\color{red}{1}} - \frac{a_{\color{red}{1}3}}{a_{33}} b_{3} \\ a_{\color{green}{2}1} = a_{\color{green}{2}1} - \frac{a_{\color{green}{2}3}}{a_{33}} a_{31} & a_{\color{green}{2}2} = a_{\color{green}{2}2} - \frac{a_{\color{green}{2}3}}{a_{33}} a_{32} & a_{\color{green}{2}3} = a_{\color{green}{2}3} - \frac{a_{\color{green}{2}3}}{a_{33}} a_{33} & a_{\color{green}{2}4} = a_{\color{green}{2}4} - \frac{a_{\color{green}{2}3}}{a_{33}} a_{34} & b_{\color{green}{2}} = b_{\color{green}{2}} - \frac{a_{\color{green}{2}3}}{a_{33}} b_{3} \\ a_{31} = a_{31} & a_{32} = a_{32} & a_{33} = a_{33} & a_{34} = a_{34} & b_{3} = b_{3} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \\ \hline a_{\color{red}{1}1} = a_{\color{red}1} - \frac{a_{\color{red}{1}2}}{a_{22}} a_{21} & a_{\color{red}{1}2} = a_{\color{red}{1}2} - \frac{a_{\color{red}{1}2}}{a_{22}} a_{22} & a_{\color{red}{1}3} = a_{\color{red}{1}3} - \frac{a_{\color{red}{1}2}}{a_{22}} a_{23} & a_{\color{red}{1}4} = a_{\color{red}{1}4} - \frac{a_{\color{red}{1}2}}{a_{22}} a_{24} & b_{\color{red}{1}} = b_{\color{red}{1}} - \frac{a_{\color{red}{1}2}}{a_{22}} b_{2} \\ a_{21} = a_{21} & a_{22} = a_{22} & a_{23} = a_{23} & a_{24} = a_{24} & b_{2} = b_{2} \\ a_{31} = a_{31} & a_{32} = a_{32} & a_{33} = a_{33} & a_{34} = a_{34} & b_{3} = b_{3} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \end{array} \end{equation*}

lo anterior puede escribirse como

\begin{align*} a_{ij} &= a_{ij} - \frac{a_{i?}}{a_{??}} a_{?j} \\ b_{i} &= b_{i} - \frac{a_{i?}}{a_{??}} b_{?} \end{align*}

para

\begin{align*} i &= 3, 2, 1 \\ &= 2, 1 \\ &= 1 \\ \end{align*}

Tercer patrón

\begin{equation*} \begin{array}{llll:l} a_{11} = a_{11} - \frac{a_{1\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}1} & a_{12} = a_{12} - \frac{a_{1\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}2} & a_{13} = a_{13} - \frac{a_{1\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}3} & a_{14} = a_{14} - \frac{a_{1\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}4} & b_{1} = b_{1} - \frac{a_{1\color{blue}{4}}}{a_{\color{blue}{44}}} b_{\color{blue}{4}} \\ a_{21} = a_{21} - \frac{a_{2\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}1} & a_{22} = a_{22} - \frac{a_{2\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}2} & a_{23} = a_{23} - \frac{a_{2\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}3} & a_{24} = a_{24} - \frac{a_{2\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}4} & b_{2} = b_{2} - \frac{a_{2\color{blue}{4}}}{a_{\color{blue}{44}}} b_{\color{blue}{4}} \\ a_{31} = a_{31} - \frac{a_{3\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}1} & a_{32} = a_{32} - \frac{a_{3\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}2} & a_{33} = a_{33} - \frac{a_{3\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}3} & a_{34} = a_{34} - \frac{a_{3\color{blue}{4}}}{a_{\color{blue}{44}}} a_{\color{blue}{4}4} & b_{3} = b_{3} - \frac{a_{3\color{blue}{4}}}{a_{\color{blue}{44}}} b_{\color{blue}{4}} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \\ \hline a_{11} = a_{11} - \frac{a_{1\color{green}{3}}}{a_{\color{green}{33}}} a_{\color{green}{3}1} & a_{12} = a_{12} - \frac{a_{1\color{green}{3}}}{a_{\color{green}{33}}} a_{\color{green}{3}2} & a_{13} = a_{13} - \frac{a_{1\color{green}{3}}}{a_{\color{green}{33}}} a_{\color{green}{3}3} & a_{14} = a_{14} - \frac{a_{1\color{green}{3}}}{a_{\color{green}{33}}} a_{\color{green}{3}4} & b_{1} = b_{1} - \frac{a_{1\color{green}{3}}}{a_{\color{green}{33}}} b_{\color{green}{3}} \\ a_{21} = a_{21} - \frac{a_{2\color{green}{3}}}{a_{\color{green}{33}}} a_{\color{green}{3}1} & a_{22} = a_{22} - \frac{a_{2\color{green}{3}}}{a_{\color{green}{33}}} a_{\color{green}{3}2} & a_{23} = a_{23} - \frac{a_{2\color{green}{3}}}{a_{\color{green}{33}}} a_{\color{green}{3}3} & a_{24} = a_{24} - \frac{a_{2\color{green}{3}}}{a_{\color{green}{33}}} a_{\color{green}{3}4} & b_{2} = b_{2} - \frac{a_{2\color{green}{3}}}{a_{\color{green}{33}}} b_{\color{green}{3}} \\ a_{31} = a_{31} & a_{32} = a_{32} & a_{33} = a_{33} & a_{34} = a_{34} & b_{3} = b_{3} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \\ \hline a_{11} = a_{11} - \frac{a_{1\color{red}{2}}}{a_{\color{red}{22}}} a_{\color{red}{2}1} & a_{12} = a_{12} - \frac{a_{1\color{red}{2}}}{a_{\color{red}{22}}} a_{\color{red}{2}2} & a_{13} = a_{13} - \frac{a_{1\color{red}{2}}}{a_{\color{red}{22}}} a_{\color{red}{2}3} & a_{14} = a_{14} - \frac{a_{1\color{red}{2}}}{a_{\color{red}{22}}} a_{\color{red}{2}4} & b_{1} = b_{1} - \frac{a_{1\color{red}{2}}}{a_{\color{red}{22}}} b_{\color{red}{2}} \\ a_{21} = a_{21} & a_{22} = a_{22} & a_{23} = a_{23} & a_{24} = a_{24} & b_{2} = b_{2} \\ a_{31} = a_{31} & a_{32} = a_{32} & a_{33} = a_{33} & a_{34} = a_{34} & b_{3} = b_{3} \\ a_{41} = a_{41} & a_{42} = a_{42} & a_{43} = a_{43} & a_{44} = a_{44} & b_{4} = b_{4} \end{array} \end{equation*}

lo anterior puede escribirse como

\begin{align*} a_{ij} &= a_{ij} - \frac{a_{ik}}{a_{kk}} a_{kj} \\ b_{i} &= b_{i} - \frac{a_{ik}}{a_{kk}} b_{k} \end{align*}

para $k = 4, 3, 2 = m, \dots, 2$

Fórmula matemática

\begin{align*} k &= m, \dots, 2 \\ & \quad i = k-1, \dots, 1 \\ & \quad \quad j = n, \dots, 1 \\ & \quad \quad \quad a_{ij} = a_{ij} - \frac{a_{ik}}{a_{kk}} a_{kj} \\ & \quad \quad b_{i} = b_{i} - \frac{a_{ik}}{a_{kk}} b_{k} \end{align*}

Seudocódigo

function eliminacion_atras(a,b)
    m, n = tamaño(a)
    for k=m to 2 do
        for i=k-1 to 1 do
            for j=n to 1 do
                a(i,j) = a(i,j) - a(i,k)*a(k,j)/a(k,k)
            end for
            b(i) = b(i) - a(i,k)*b(k)/a(k,k)
        end for
    end for
end function

otra alternativa para reducir tiempo de cálculo

function eliminacion_atras(a,b)
    m, n = tamaño(a)
    for k=m to 2 do
        for i=k-1 to 1 do
            factor = a(i,k)/a(k,k)
            for j=n to 1 do
                a(i,j) = a(i,j) - a(i,k)*a(k,j)/a(k,k)
            end for
            b(i) = b(i) - a(i,k)*b(k)/a(k,k)
        end for
    end for
end function

Implementación eliminación hacia atrás



In [1]:

    
import numpy as np

def eliminacion_atras(A,B):
    a = np.copy(A)
    b = np.copy(B)
    m, n = a.shape
    for k in range(m-1,-1,-1):
        for i in range(k-1,-1,-1):
            factor = a[i,k]/a[k,k]
            for j in range(n-1,-1,-1):
                a[i,j] = a[i,j] - factor*a[k,j]
            b[i,0] = b[i,0] - factor*b[k,0]
    return a,b



In [2]:

    
A = np.array([[3,-0.1,-0.2],[0.1,7,-0.3],[0.3,-0.2,10]])
print(A)









    



[[  3.   -0.1  -0.2]
 [  0.1   7.   -0.3]
 [  0.3  -0.2  10. ]]



In [3]:

    
B = np.array([7.85,-19.3,71.4]).reshape((3,1))
print(B)









    



[[  7.85]
 [-19.3 ]
 [ 71.4 ]]



In [4]:

    
eliminacion_atras(A,B)[0]









    Out[4]:





array([[  3.00762082,   0.        ,   0.        ],
       [  0.109     ,   6.994     ,   0.        ],
       [  0.3       ,  -0.2       ,  10.        ]])



In [5]:

    
eliminacion_atras(A,B)[1]









    Out[5]:





array([[  9.02286245],
       [-17.158     ],
       [ 71.4       ]])

Sustitución hacia adelante

Resolver el sistema de ecuaciones

Incógnita $x_{1}$

\begin{equation*} x_{1} = \frac{9.022862}{3.007621} = 3 \end{equation*}

Incógnita $x_{2}$

\begin{equation*} x_{2} = \frac{-17.158 - (0.109)(3)}{6.994} = -2.5 \end{equation*}

Incógnita $x_{1}$

\begin{equation*} x_{1} = \frac{71.4 - (0.3)(3) - (-0.2)(-2.5)}{10} = 7 \end{equation*}

Algoritmo de cálculo

\begin{alignedat}{5} a_{11} x_{1} & & & & {}={} b_{1} \\ a_{21} x_{1} & {}+{} a_{22} x_{2} & & & {}={} b_{2} \\ a_{31} x_{1} & {}+{} a_{32} x_{2} & {}+{} a_{33} x_{3} & & {}={} b_{3} \\ a_{41} x_{1} & {}+{} a_{42} x_{2} & {}+{} a_{43} x_{3} & {}+{} a_{44} x_{4} & {}={} b_{4} \end{alignedat}

Incognita $x_{1}$

\begin{equation*} x_{1} = \frac{b_{1}}{a_{11}} \end{equation*}

Incognita $x_{2}$

\begin{equation*} x_{2} = \frac{b_{2} - a_{21} x_{1}}{a_{22}} \end{equation*}

Incognita $x_{3}$

\begin{equation*} x_{3} = \frac{b_{3} - a_{31} x_{1} - a_{32} x_{2} }{a_{33}} \end{equation*}

Incognita $x_{4}$

\begin{equation*} x_{4} = \frac{b_{4} - a_{41} x_{1} - a_{42} x_{2} - a_{43} x_{3} }{a_{44}} \end{equation*}

Patrón de cálculo

Primer patrón

\begin{align*} x_{\color{blue}{1}} &= \frac{b_{\color{blue}{1}}}{a_{\color{blue}{11}}} \\ x_{2} &= \frac{b_{2} - a_{21} x_{1}}{a_{22}} \\ x_{3} &= \frac{b_{3} - a_{31} x_{1} - a_{32} x_{2} }{a_{33}} \\ x_{4} &= \frac{b_{4} - a_{41} x_{1} - a_{42} x_{2} - a_{43} x_{3} }{a_{44}} \end{align*}

lo anterior puede escribirse como

\begin{equation*} x_{1} = \frac{b_{1}}{a_{11}} \end{equation*}

Segundo patrón

\begin{align*} x_{1} &= \frac{b_{1}}{a_{11}} \\ x_{2} &= \frac{b_{2} - a_{2\color{blue}{1}} x_{\color{blue}{1}}}{a_{22}} \\ x_{3} &= \frac{b_{3} - a_{3\color{blue}{1}} x_{\color{blue}{1}} - a_{3\color{green}{2}} x_{\color{green}{2}} }{a_{33}} \\ x_{4} &= \frac{b_{4} - a_{4\color{blue}{1}} x_{\color{blue}{1}} - a_{4\color{green}{2}} x_{\color{green}{2}} - a_{4\color{red}{3}} x_{\color{red}{3}} }{a_{44}} \end{align*}

lo anterior puede escribirse como

\begin{equation*} x_{?} = \frac{b_{?} - \sum_{j}^{n} a_{?j} x_{j}}{a_{??}} \end{equation*}

para

\begin{align*} j &= 1 \\ &= 1, 2 \\ &= 1, 2, 3 \end{align*}

Tercer patrón

\begin{align*} x_{1} &= \frac{b_{1}}{a_{11}} \\ x_{\color{blue}{2}} &= \frac{b_{\color{blue}{2}} - a_{\color{blue}{2}1} x_{1}}{a_{\color{blue}{22}}} \\ x_{\color{green}{3}} &= \frac{b_{\color{green}{3}} - a_{\color{green}{3}1} x_{1} - a_{\color{green}{3}2} x_{2} }{a_{\color{green}{33}}} \\ x_{\color{red}{4}} &= \frac{b_{\color{red}{4}} - a_{\color{red}{4}1} x_{1} - a_{\color{red}{4}2} x_{2} - a_{\color{red}{4}1} x_{4} }{a_{\color{red}{44}}} \end{align*}

lo anterior puede escribirse como

\begin{equation*} x_{i} = \frac{b_{i} - \sum_{j}^{n} a_{ij} x_{j}}{a_{ii}} \end{equation*}

para $i = 2, 3, 4 = 2, \dots, m$

Fórmula matemática

\begin{align*} x_{1} &= \frac{b_{1}}{a_{11}} \\ i &= 2, \dots, m \\ & \quad x_{i} = \frac{b_{i} - \sum_{j=1}^{i-1} a_{ij} x_{j}}{a_{ii}} \end{align*}

Seudocódigo

function sustitucion_adelante(a,b)
    m, n = tamaño(a)
    x(1) = b(1)/a(1,1)
    for i=2 to m do
        sumatoria = b(i)
        for j=1 to i-1 do
            sumatoria = sumatoria - a(i,j)*x(j)
        end for
        x(i) = sumatoria / a(i,i)
    end for
    return x
end function

Implementación sustitución hacia adelante



In [6]:

    
import numpy as np

def sustitucion_adelante(a,b):
    m, n = a.shape
    x = np.zeros(b.shape)
    x[0,0] = b[0,0]/a[0,0]
    for i in range(1,m):
        sumatoria = b[i,0]
        for j in range(0,i):
            sumatoria = sumatoria - a[i,j]*x[j,0]
        x[i,0] = sumatoria/a[i,i]
    return x



In [7]:

    
A = np.array([[3.007621,0,0],
              [0.109,6.994,0],
              [0.3,-0.2,10]])
print(A)









    



[[  3.007621   0.         0.      ]
 [  0.109      6.994      0.      ]
 [  0.3       -0.2       10.      ]]



In [8]:

    
B = np.array([[9.022862],
              [-17.158],
              [71.4]])
print(B)









    



[[  9.022862]
 [-17.158   ]
 [ 71.4     ]]



In [9]:

    
sustitucion_adelante(A,B)









    Out[9]:





array([[ 2.99999967],
       [-2.49999999],
       [ 7.00000001]])



In [10]:

    
#revisando el resultado
X = np.linalg.solve(A, B)
print(X)
#revisando la solución
np.allclose(np.dot(A, X), B)









    



[[ 2.99999967]
 [-2.49999999]
 [ 7.00000001]]






    Out[10]:





True



In [11]:

    
A = np.array([[1,0,0],[2,3,0],[4,5,6]])
print(A)









    



[[1 0 0]
 [2 3 0]
 [4 5 6]]



In [12]:

    
B = np.array([[7],[8],[9]])
print(B)









    



[[7]
 [8]
 [9]]



In [13]:

    
sustitucion_adelante(A,B)









    Out[13]:





array([[ 7. ],
       [-2. ],
       [-1.5]])



In [14]:

    
#revisando el resultado
X = np.linalg.solve(A, B)
print(X)
#revisando la solución
np.allclose(np.dot(A, X), B)









    



[[ 7. ]
 [-2. ]
 [-1.5]]






    Out[14]:





True

Implementación eliminación de Gauss



In [15]:

    
def eliminacion_gauss(a,b):
    A, B = eliminacion_atras(a,b)
    x = sustitucion_adelante(A,B)
    print(x)



In [16]:

    
C = np.array([[4,-1,1],
              [2,5,2],
              [1,2,4]],float)
D = np.array([[8],
              [3],
              [11]],float)



In [17]:

    
eliminacion_gauss(C,D)









    



[[ 1.]
 [-1.]
 [ 3.]]



In [18]:

    
#revisando el resultado
solucion = np.linalg.solve(C, D)
print(solucion)
#revisando la solución
np.allclose(np.dot(C, solucion), D)









    



[[ 1.]
 [-1.]
 [ 3.]]






    Out[18]:





True



In [25]:

    
E = np.array([[1,5,-1,1,-1],
              [2,2,4,-1,1],
              [3,12,-3,-2,3],
              [4,10,-2,4,-5],
              [16,-10,6,-1,-1]],float)
F = np.array([[2],
              [4],
              [8],
              [16],
              [32]],float)



In [26]:

    
eliminacion_gauss(E,F)









    



[[ -0.125     ]
 [  1.42105263]
 [ -1.19078947]
 [-30.76315789]
 [-24.59210526]]



In [27]:

    
#revisando el resultado
solucion = np.linalg.solve(E, F)
print(solucion)
#revisando la solución
np.allclose(np.dot(E, solucion), F)









    



[[ -0.125     ]
 [  1.42105263]
 [ -1.19078947]
 [-30.76315789]
 [-24.59210526]]






    Out[27]:





True