un fissema lineal de la forma a x = b {\ displaystyle ax = b}
con una stima non identica x (0) {\ displaystyle x ^ {(0)}}
está dado por a =, b = yx (0) =. {\ displaystyle a = {\ begin {bmatrix} 2 & 1 \\ 5 & 7 \\- End {BMatrix}}, \ b = {\ begin {bmatrix} 11 \\ 13 \\\ fine {bmatrix}} \ quad {\ text {y}}} quad x ^ {(0)} = {\ begin {bmatrix} 1 \\ 1 \\\ fine {bmatrix}}.}
usamos la ecuación x (k + 1) = D – 1 (b – r x (k)) {\ displaystyle x ^ {(k + 1)} = D ^ {- 1} (B-RX ^ {(K)})}
, DESCRITA anteriormente, para prelier x {\ displaystyle x}
. Primero, Reiscricimos 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 que r = l + u {\ displaystyle r = l + u}
donde l {\ displaystyle l }
y u {\ displaystyle u}
figlio las Partes inferiori y superiore de a {\ displaystyle a}
. De los Valores Conocidos. D – 1 =, l = y u =. {\ displaystyle d ^ {- 1} = {\ begin {bmatrix} 1/2 & 0 \\ 0 & 1/7 \\\ fine {bmatrix}}, \ l = {\ begin {bmatrix} 0 & 0 \\ 5 & 0 \\ \ end {bmatrix}}} quad {\ text {y}} \ quad u = {\ begin {bmatrix} 0 & 1 \\ 0 & 0 \\\ 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}} \ \ sinistra \ {{\ begin {bmatrix} 0 & 0 \\ – 5 & 0 \\\ End {BMatrix }} + {\ begin {bmatrix} 0 & -1 \\ 0 & 0 \\\ End {BMatrix}} \ Destra \} = {\ begin {bmatrix} 0 & -0.5 \\ – 0.714 & 0 \\\ End {BMatrix}}. }
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 \\\- fine {bmatrix}}.}
Con t y c calcolamas, stimaremos x {\ displaystyle x}
como x (1) = t x (0) + c {\ displaystyle x ^ {\ displaystyle x ^ {(1)} = tx ^ {(0)} + c }
: x (1) = + =. {\ displaystyle x ^ {(1)} = {\ begin {bmatrix} 0 & -0.5 \\ – 0.714 & 0 \ \\ fine {bmatrix}} {\ begin {bmatrix} 1 \\ 1 \\\ end {bmatrix}} + {\ begin {bmatrix} 5.5 \\ 1.857 \\\ win {bmatrix}} = {\ begin {bmatrix } 5.0 \\ 1.143 \\\ end {bmatrix}}.}
siguientes iteraciones.
x (2) = + =. {\ displaystyle x ^ {(2)} = {\ begin {bmatrix} 0 & -0.5 \\ – 0.714 & 0 \ \\ end {bmatrix}} {\ begin} {bmatrix} 5.0 \\ 1.143 \\\ end {bmatrix}} + {\ begin {bmatrix} 5.5 \\ 1.857 \\\ end {bmatrix}} = {\ Begin {BMatrix } 4.929 \\ – 1.713 \\\ fine {bmatrix}}.}
