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If we have a lower triangular matrix $$A=\left(\begin{array}{rrrrr}a_{1,1}&0&0&\cdots&0\\a_{2,1}&a_{2,2}&0&\cdots&0\\a_{3,1} &1_{3,2}&a_{3,3}&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\a_{n,1}&a_{n,2}&a_{n,3}&\cdots&a_{n,n}\end{array}\right)$$ does there exist a formula for a certain element of $A^{-1}$ in terms of its coordinates, the elements of $A$, and $\det A$?

Given such a matrix of any particular size, we can calculate each term of the inverse matrix by expanding the multiplication and solving. For instance, if $A$ is a $3\rm x 3$ matrix, then we have $$A^{-1}=\left(\begin{array}{lll} \frac 1{a_{1,1}}&0&0\\ -a_{2,1}\over a_{1,1}a_{2,2}&\frac 1{a_{2,2}}&0\\ a_{3,2}a_{2,1}-a_{3,1}a_{2,2}\over a_{1,1}a_{2,2}a_{3,3}&-a_{3,2}\over a_{2,2}a_{3,3}&\frac 1{a_{3,3}} \end{array}\right).$$ We can, by similar methods, calculate exact formulae for any specific value of $n$. These formulae seem tantalizingly as if there is some general formula that holds for all $n$ for each entry, but I am unable to find one. Does such a formula exist?

Note 1: I am aware of a few other questions on this subject, but none of them seem to answer this; however, I know very little linear algebra I may have just failed to understand them.

Note 2: Spellcheck is telling me that "formulae" is not a word. This annoys me. However, it least it also says that "spellcheck" is not a word. Sorry. I'll stop now.

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  • $\begingroup$ What is $b_{n,k}$? $\endgroup$ – Robert Lewis Mar 14 '15 at 18:11
  • $\begingroup$ @RobertLewis The element of $A^{-1}$ in the $n$th row, $k$th column. Editing question to clarify. $\endgroup$ – Laertes Mar 14 '15 at 18:29
  • $\begingroup$ I see now. Thanks! $\endgroup$ – Robert Lewis Mar 14 '15 at 18:31

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