Método de Jacobi

Un Sistema Lineal de la forma A x = B {\ displaystyle Ax = B}

<00bf135406 "> {\ displaystyle ax = B}

Contacte estime incial x (0) {\ displaystyle x ^ {(0)}}

${\ displaystyle x ^ {(0)} }$

está dado por a =, b = yx (0) =. {\ displaystyle a = {\ begin {bmatrix} 2 & 1 \\ 5 & 7 \\\ fin {bmatrix}}, \ b = {\ begin {bmatrix} 11 \\ 13 \\\ fin {bmatrix}} \ quad {\ texte {y}} \ quad x ^ {(0)} = {\ commencez {bmatrix} 1 \\ 1 \\\ fin {bmatrix}}.}

${\ displaystyle a = {\ commencez {bmatrix} 21 \\ 57 \\\ fin {bmatrix}}, \ b = {\ commencer {bmatrix} 11 \\ 13 \\\ fin {bmatrix}} \ quad {\ texte {y}} \ quad x ^ {(0)} = {\ commencez {bmatrix} 1 \\ 1 \\\ fin { bmatrix}}.}$

usamos la Ecuación x (k + 1) = D – 1 (B – R x (k)) {\ displaystyle x ^ {(K + 1)} = D ^ {- 1} (b-rx ^ {(k)})}

${\ displaystyle x ^ {(k + 1)} = D ^ { -1} (b-rx ^ {(k)})}$

, descrita anteriormente, para estimation x {\ displaystyle x}

$x$

. PRIMERO, REESCRIMOS LA ECUACIÓN DE UNA MANAGERA MÁS CONDITIONNE D – 1 (B – R X (K)) = T x (k) + C {\ DisplayStyle D ^ {- 1} (B-RX ^ {(K)}) = Tx ^ {(k)} + c}

${\ displaystyle d ^ {- 1} (b-rx ^ {(k)}) = tx ^ {(k)} + C}$

, Donde T = – D – 1 R {\ displaystyle t = -D ^ {- 1} r}

${\ displaystyle t = -D ^ {- 1} r}$

y c = d – 1 b {\ displaystyle c = d ^ {- 1} b}

${\ displaystyle C = D ^ {- 1} B}$

. Vea que r = l + u {\ displaystyle r = l + u}

${\ displaystyle r = l + u}$

donde l {\ displaystyle l }

$l$

y u {\ displaystyle u}

<3013875aec "> u

fils las Partes inférieur y supérieur de a {\ displaystyle a}

$A$

. de los valores conocidos. D – 1 =, l = y u =. {\ displaystyle d ^ {- 1} = {\ commencez {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ fin {bmatrix}}, \ l = {\ commencez {bmatrix} 0 & 0 \\ 5 & 0 \\ \ fin {bmatrix}} \ quad {\ texte {y}} \ quad u = {\ commencez {bmatrix} 0 & 1 \\ 0 & 0 \\\ fin {bmatrix}}.}

${\ displaystyle d ^ {- 1} = {\ begin {bmatrix} 1/20 \\ 01/7 \\\ fin {bmatrix}}, \ l = {\ begin {bmatrix} 00 \\ 50 \\\ fin {bmatrix}} \ quad {\ texte {y}} \ quad u = {\ commencez {bmatrix} 01 \\ 00 \\\ fin {bmatrix}}.}$

déterminé t = – d – 1 (l + u) {\ displaystyle t = -d ^ {- 1} (l + U)}

« a904590e96″> {\ displaystyle t = -d ^ {- 1} (l + u)}

como t = {+ } =. {\ displaystyle t = {\ commencez {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ fin { BMATRIX}} \ GAUCHE \ {{\ COMMENTER {BMATRIX} 0 & 0 \\ – 5 & 0 \\\ fin {bmatrix }}} + {\ begin {bmatrix} 0 & -1 \\ 0 & 0 \\\ fin {bmatrix}} \ droite \} = {\ begin {bmatrix} 0 & -0.5 \\ – 0.714 & 0 \\\ fin {bmatrix}}. }

${\ displaystyle t = {\ begin {bmatrix} 1/20 \\\ 01/7 \\\ fin {bmatrix}} \ gast \ {{\ begin {bmatrix} 00 00 \\ - 50 \\\ fin {bmatrix}} + {\ begin {bmatrix} 0-1 \\ 00 \\\ fin {bmatrix}} \ droite \} = {\ begin {bmatrix} 0-0.5 \\ - 0.7140 \\\ fin {bmatrix}}.}$

C es encontrada como

c = =. {\ displaystyle c = {\ commencez {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ fin { bmatrix}} {\ begin {bmatrix} 11 \\ 13 \\\ fin {bmatrix}} = {\ commencent {bmatrix} 5.5 \\ 1.857 \\\ fin {bmatrix}}.}

69cc17d0a1 « > {\ displaystyle c = {\ commencements {bmatrix} 1/20 \\ 01/7 \\\ fin {bmatrix}} {\ begin {bmatrix} 11 \\ 13 \\\ fin {bmatrix}} = { \ begin {bmatrix} 5.5 \\ 1.857 \\\ fin {bmatrix}}.}

content {\ displaystyle x}

$x$

Côme x (1) = t x (0) + c {\ displaystyle x ^ {(1)} = tx ^ {(0)} + c }

${\ displaystyle x ^ {(1)} = tx ^ {(0)} + c}$

: x (1) = + =. {\ displaystyle x ^ {(1)} = {\ commencez {bmatrix} 0 & -0.5 \\ – 0.714 & 0 \ \\ fin {bmatrix}} {\ begin {bmatrix} 1 \\ 1 \\\ fin {bmatrix}} + {\ begin {bmatrix} 5.5 \\ 1.857 \\\ fin {bmatrix}} = {\ begin {bmatrix } 5.0 \\ \\ \\ 1.143 \\\ fin {bmatrix}}.}

${\ displaystyle x ^ {(1)} = {\ commencez {bmatrix} 0-0.5 \\ - 0.7140 \\\ fin {bmatrix}} {\ begin {bmatrix} 1 \\ 1 \\\ fin {bmatrix}} + {\ begin {bmatrix} 5.5 \\ 1.857 \\\ fin {bmatrix}} = {\ commencez { bmatrix} 5.0 \\ 1.143 \\\ fin {bmatrix}}.}$

siguities itéraciones.

x (2) = + = +. {\ displaystyle x ^ {(2)} = {\ begin {bmatrix} 0 & -0.5 \\ – 0.714 & 0 \ \\ fin {bmatrix}} {\ begin {bmatrix} 5.0 \\ 1.143 \\\ fin {bmatrix}} + {\ commencez {bmatrix} 5.5 \\ 1.857 \\\ fin {bmatrix}} = {\ commencements {bmatrix } 4.929 \\ – 1.713 \\\ fin {bmatrix}}.}

${\ displaystyle x ^ {(2)} = {\ commencez {bmatrix} 0-0.5 \\ - 0.7140 \\\ fin {bmatrix}} {\ begin {bmatrix} 5.0 \\ 1.143 \\\ fin {bmatrix}} + {\ commencez {bmatrix} 5.5 \\ 1.857 \\\ fin {bmatrix}} = {\ commencez {bmatrix} 4.929 \\ - 1.713 \\\ fin {bmatrix}}.}$

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