Interestingly, no one tried to solve this analytically and get: 3 x1 = 4 / 38 x2 = -9/38 x3 = 25/38 Seems those values could be obtained in this form using Jacobi algorithm without going to decimal places.
nice work done bro, but take note when you were solving the 2nd iteration for x2 instead of 1/6[0-3(1/3)-2(4/7) ] you made it 1/6[0-3(1/3)-2(-4/7)]. the 4/7 was supposed to be positive not negative . even though your answer was correct
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wowww.. this is a quick method... actually made my life easier.. :)
Thank you
Interestingly, no one tried to solve this analytically and get:
3 x1 = 4 / 38
x2 = -9/38
x3 = 25/38
Seems those values could be obtained in this form using Jacobi algorithm without going to decimal places.
buh please how will you determine infinite-norm from the Jacobi method given initial conditions like, x(k)-x(k+1) for k=1,2,3,...n
We will make videos on that for you please
nice work done bro, but take note
when you were solving the 2nd iteration for x2 instead of 1/6[0-3(1/3)-2(4/7) ]
you made it 1/6[0-3(1/3)-2(-4/7)]. the 4/7 was supposed to be positive not negative .
even though your answer was correct
Thank you Bernard, guess it was a typo
@@ReindolfBoadu yh sure
please where can i get more questions
Use this link please
www.sanfoundry.com/matrix-inversion-questions-answers-jacobis-iteration-method/
two words for you. Thank you. No thank you. No Really thank you.
You are welcome Sir
Charlie your accent 😂😂 - I’m Nigerian
the Third iteration i didnt get the same answer with you
the 2nd iteration of X2 i got 1/42
hey that's my name
Yeah 😀
@@ReindolfBoadu you earned a subscriber for making a method out of my name!