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 Author Comment/Response Bill Simpson 09/07/12 9:37pm From In[11]:= B={{1.1161, 0.1254, 0.1397, 0.149, 1.5471}, {0.1582, 1.1675, 0.1768, 0.1871, 1.4471}, {0.1968, 0.2071, 1.2168, 0.2271, 1.7471}, {0.2368, 0.2471, 0.2568, 1.2671, 1.8471}) In[12]:=(*We must process all four columns*) For[column=1,column≤4,column++, (*Find the pivot row*) (*Begin by assuming the first row is the pivot row*) pivotrow=1; For[row=2,row≤4,row++, If[B[[row,column]]>B[[pivotrow,column]], (*element is larger, revise pivot row assumption*) pivotrow=row] ]; (*Now we know the pivot row*) (*Process all four rows*) For[row=1,row≤4,row++, If[row≠pivotrow, (*then we subtract scaled pivot row from row*) B[[row]]=B[[row]]-B[[row,column]]/B[[pivotrow,column]]*B[[pivotrow]], (*else we scale the row to have a leading 1*) B[[row]]=B[[row]]/B[[row,column]] ] ]; ]; Print["B=",MatrixForm[B]]; From In[19]:=B= {{1.0000, 1.3877*^-17, 0.0000, -1.3877*^-17, 1.0542}, {0.0000, 1.0000, 0.0000, 0.0000, 0.8057}, {0.0000, 0.0000, 1.0000, -2.7755*^-17, 0.9584}, {0.0000, 0.0000, 0.0000, 1.0000, 0.9093}} Study this and test this very carefully to make certain that I have made no mistakes in this. But there is some chance this is at least partly correct because the result seems to match In[31]:= LinearSolve[A,RHS] Out[31]= {1.05427,0.80576,0.958446,0.909333} URL: ,

 Subject (listing for 'Gaussian elimination methods') Author Date Posted Gaussian elimination methods Tatjana 09/07/12 09:42am Re: Gaussian elimination methods Bill Simpson 09/07/12 9:37pm Re: Gaussian elimination methods Bill Simpson 09/07/12 11:53pm Re: Re: Gaussian elimination methods Tatjana 09/08/12 1:19pm Re: Re: Re: Gaussian elimination methods Bill Simpson 09/08/12 5:57pm
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