Solving for the coefficients in the "A" lower triangular matrix for the form Ax=b

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GS76
GS76 2015 年 12 月 16 日
回答済み: John D'Errico 2015 年 12 月 16 日
Hi all,
Could you assist me with a way for solving for the coefficients in the "A" lower triangular matrix for the form Ax=b?
Please see the attached spread sheet with the known and unknown variables.
Thank you.
Regards,
Gary
  1 件のコメント
Renato Agurto
Renato Agurto 2015 年 12 月 16 日
編集済み: Renato Agurto 2015 年 12 月 16 日
I think you can't. You have 10 independent equations. the 1st with 1 (a11) variable. the 2nd with 2 (a21, a22), the 3rd with 3(a31, a32, a33), etc.
If there are no more restricctions for the contents of A, then you can only calculate a11.

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John D'Errico
John D'Errico 2015 年 12 月 16 日
Not possible. Renato is correct, in that you have essentially only 10 pieces of information, in the form of 10 linear equations.
However, you have 55 unknowns. You can pick some subset 45 of those unknowns, and set them to zero (or any value you choose) and then solve for the other 10 unknowns. Not ANY arbitrary subset of course. Thus...
The single element from row 1 is easily computed.
Next, you can choose any one element from row 2 to set to any chosen value. Then the other element is given.
Repeat for each row, setting n-1 elements on the nth row to some arbitrary values, then compute the one element that remains.
You can do no better than this though. Sorry, but mathematics is unforgiving.

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