Método de Jacobi (Română)

ONU SISTEMA LINEAL DE LA FORMA A X = B {\ DisplayStyle Ax = B}

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

está dado por a =, b = yx (0) =. {\ Displaystyle A = {\ begin {bmatrix} 2 & 1 \\ 5 & 7 \\\ end {bmatrix}}, \ b = {\ începe {bmattrix} 11 \\ 13 \\\ end {bmatrix}} quad {\ text {y}} \ quad x ^ {(0)} = {\ Begin {bmatrix} 1 \\ 1 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ {\ începe {bmatrix} 11 \\ 13 \\\ end {bmatrix}} \ quad {\ text {y}} \ quad x ^ {(0)} = {\ begin {bmatrix} 1 \\ 1 \\\ end { bmattrix}}.}

USAMOS LA Ecuación X (K + 1) = D – 1 (B – R X (K)) {\ DisplayStyle x ^ {(K + 1)} = D ^ {- 1} (b-Rx ^ {(k)})}

, descrieri anteriormente, parasusar x {\ displaystyle 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}

, donde T = – D – 1 R {\ displaystyle T = -D ^ {- 1} R}

y C = D – 1 b {\ displaystyle C = D ^ {- 1} b}

. VEA QuR = L + U {\ DisplayStyle R = L + U}

Donde L {\ DisplayStyle L }

y u {\ displaystyle u}

fiul las Parts Inferior Y Superior De A {\ Afișări A}

. De los valores conocidos. D – 1 =, l = y u =. {\ DisplayStyle D ^ {- 1} = {\ Begin {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ end {bmatrix}}, \ L = {\ begin {bmatrix} 0 & 0 \\ 5 & 0 \\ \ capătul {BMATRIX}} \ quad {\ text {y}} quad u = {\ Begin {bmattrix} 0 & 1 \\ 0 & 0 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\/7 \\\ end {bmatrix}}, \ L = {\ begin {bmatrix} 00 \\ 50 \\\ end {bmatrix}} \ quad {\ text {y}} \ quad U = {\ begin {bmatrix} 01 \\ 00 \\\ end {bmatrix}}}

determinamos T = -. D – 1 (L + U) {\ displaystyle T = -D ^ {- 1} (L + U)}

como t = {+ } =. {\ Displaystyle T = {\ begin {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ end { bmatrix}} \ stângă \ {{\ begin {bmatrix} 0 & 0 \\ – 5 & 0 \\\ end {bmatrix }} + {\ begin {bmatrix} 0 & -1 \\ 0 & 0 \\\ end {bmatrix}} \ dreapta \} = {\ Begin {bmattrix} 0 &

0 \\\ end {bmattrix}}. }

C es encontrada como

c = = =. {\ Displaystyle C = {\ begin {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ end { bmatrix}} {\ begin {bmatrix} 11 \\ 13 \\\ end {bmatrix}} = {\ begin {bmatrix} 5.5 \\ 1.857 \\\ end {bmatrix}}.}

con T calculadas y C, estimaremos x {\ displaystyle x}

como x (1) = t x (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 \\\ end {bmatrix}} + {\ begin {bmatrix} 5.5 \\ 1.857 \\\ end {bmatrix}} = {\ begin {bmatrix } 5.0 \\ 1.143 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ {} \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ bmatrix} 5.0 \\ 1,143 \\\ end {bmatrix}}.}