Método de jacobi

un sistema de sistema lineal de la forma a x = b {\ displaystyle ax = b}

${\ displaystyle hacha = b}$

con unha estimación inicial x (0) {\ displaystyle x ^ {(0)}}

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

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

${\ displaystyle a = {\ begin {bmatrix} 21 \\ 57 \\\ {bmatrix}}, \ b = {\ begin {bmatrix} 11 \\ 13 \\\\ {bmatrix}} \ quad {\ texto {y}} \ quad x ^ {(0)} = {\ begin {bmatrix} 1 \\ 1 \\\ { bmatrix}}.}$

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 estimar x {\ displaystyle x}

$x$

. Primero, reescribimos la ecuación de una manera más conveniente 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}

$u$

son las Partes inferiores y superior de a {\ displaystyle a}

$a$

. dos valores conocidos. D – 1 =, l = y u =. {\ displaystyle d ^ {- 1} = {\ begin {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ {{bmatrix}}, \ l = {\ begin {bmatrix} 0 & 0 \\ 5 & 0 \\ \ END {BMATRIX}} \ quad {\ TEXT {Y}} \ quad u = {\ begin {bmatrix} 0 & 1 \\ 0 & 0 \\\\ end {bmatrix}}.}

${\ displaystyle d ^ {- 1} = {\ begin {bmatrix} 1/20 01/7 \\\ end {bmatrix}}, \ l = {\ begin {bmatrix} 00 \\ 50 \\\\ {bmatrix}} \ quad {\ texto {}}} \ quad u = {begin {bmatrix} 01 \\ 00 \\\\ {bmatrix}}.}$

Determinamos t = – D – 1 (L + U) {\ displaystyle t = -d ^ {- 1} (l + U)}

${\ displaystyle t = -d ^ {- 1} (l + u)}$

como t = {+ + } =. {\ displaystyle t = {\ begin {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ { bmatrix}} \ left \ {{\ begin {bmatrix} 0 & 0 \\ – 5 & 0 \\\ {end {bmatrix }} + {\ begin {bmatrix} 0 & -1 \\ 0 & 0 \\\\ {bmatrix}} \ rectual \} = {\ begin {bmatrix} 0 & -0.5 \\ – 0.714 & 0 \\\ {bmatrix}}. }

${\ displaystyle t = {\ begin {bmatrix} 1/20 \\ 01/7 \\\ {{bmatrix}} \ left \ {{\ begin {bmatrix} 00 \\ - 50 \\\\ {{bmatrix}} + {\ begin {bmatrix} 0-1 \\ 00 \\\\ {bmatrix}} \ right \} = {\ begin {bmatrix} 0-0.5 \\ - 0,7140 \\\ {bmatrix}}}$

c es encontro como

c = = =. {\ displaystyle c = {\ begin {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ { bmatrix}} {\ begin {bmatrix} 11 \\ 13 \\\ {bmatrix}} = {\ begin {bmatrix} 5,5 \\ 1.857 \\\\ {bmatrix}}}

con t y c calculadas, estimaremos x {\ displaystyle x}

$x$

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

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

: x (1) = + =. {\ displaystyle x ^ {(1)} = {\ begin {bmatrix} 0 & -0.5 \\ – 0.714 & 0 \ \\ end {bmatrix}} {\ begin {bmatrix} 1 \\ 1 \\\ {bmatrix}} + {\ begin {bmatrix} 5,5 \\ 1.857 \\\\ {bmatrix}} = {{begin {bmatrix } 5.0 \\ 1.143 \\\ {bmatrix}}.}

${\ displaystyle x ^ {(1)} = {\ begin {bmatrix} 0-0.5 \\ - 0.7140 \\\ end {bmatrix}} {\ begin {bmatrix} 1 \\ 1 \\\ {bmatrix}} + {\ begin {bmatrix} 5,5 \\ 1.857 \\\ {bmatrix}} = {{begin { bmatrix} 5.0 \\ 1.143 \\\\ {bmatrix}}.}$

siguientes iteraciones.

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

${\ displaystyle x ^ {(2)} = {\ begin {bmatrix} 0-0.5 0,7140 \\\\ {bmatrix}} {\ begin {bmatrix} 5.0 \\ 1.143 \\\ {{bmatrix}} + {\ begin {bmatrix} 5.5 \\ 1.857 \\\ {bmatrix}} = {\ begin {BMATRIX} 4.929 \\ - 1.713 \\\\ {bmatrix}}}$

Método de jacobi (Galego)

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