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(Solved):           Me remaining A Consider the question: Apply five iter ...



Me remaining A
Consider the question:
Apply five iterations of the Gauss-Seidel method, with initial approximation
(0)
to app

 

B.
xm=[0.25;0.25]
b=[6;-4]
L=[0 0;1 0]
D=[7 0;0 -5]
U=[0 -1;0 0]
bdash=iny (L+D) *b;
B=inv (L+D) *U;
for n=1:5
xm=bdash-B*xm

 

??.
xm=[0.25;0.25]
b=[6;-4]
L=[0 0:1 0]
D=[8 0;0 -6]
??
U=[0 -1:0 0]
bdash=iny (L+D) *b;
B=iny (L+D) *U;
for n=1:5
xm=bdash-B

 

D.
xm=[0.2;0.2]
b=[6;-4]
L=[0 0:1 0]
D=[8 0;0 -6]
U=[0 -1;0 0]
bdash=iny (L+D) *b;
B=inv (L+D) *U;
for n=1:5
xm=bdash-B*xm
en

 

OC.
xm=[0.25;0.25]
b=[6;-4]
L=[00:10]
D=[8 0;0 -6]
U=[0 -1;0 0]
A
bdash=iny (L+D) *b;
B=iny (L+D) *U;
for n=1:3
xm=bdash-B*xm

 

Me remaining A Consider the question: Apply five iterations of the Gauss-Seidel method, with initial approximation (0) to approximate the solution of the system: (0) 1 0.25 = 12 9 8.21 - 22 6 11 – 6x2 - Which of the following Octave codes can be used to solve the above question? A xm=[0.25:0.25] B. xm=[0.25;0.25] b=[6;-4] L=[0 0;1 0] D=[7 0;0 -5] U=[0 -1;0 0] bdash=iny (L+D) *b; B=inv (L+D) *U; for n=1:5 xm=bdash-B*xm endfor ??. xm=[0.25;0.25] b=[6;-4] L=[0 0:1 0] D=[8 0;0 -6] ?? U=[0 -1:0 0] bdash=iny (L+D) *b; B=iny (L+D) *U; for n=1:5 xm=bdash-B*xm endfor D. xm=[0.2;0.2] b=[6;-4] L=[0 0:1 0] D=[8 0;0 -6] U=[0 -1;0 0] bdash=iny (L+D) *b; B=inv (L+D) *U; for n=1:5 xm=bdash-B*xm endfor OC. xm=[0.25;0.25] b=[6;-4] L=[00:10] D=[8 0;0 -6] U=[0 -1;0 0] A bdash=iny (L+D) *b; B=iny (L+D) *U; for n=1:3 xm=bdash-B*xm endfor


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Hello Dear Student.. Here correct Option : (A) EXPLANATION : Here we have to take 5 Iterations but in option C there are only 3 iterations from n=1 to 3. Hence Option (C) is eliminated. We ha
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