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I am dealing with the following matrix equation.

$$AB= I.$$

All are square matrices. $A$ is a known tridiagonal matrix, $I$ is identity matrix. Since $B$ is unusually large, I wonder if it is possible to find out the first row of $B$ matrix without evaluating the whole one?

Please note that, neither $A$ nor $B$ is hermitian.

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    Note that if $AB = I$, then also $BA = I$, and if $A$ is nice, that makes computing the first row of $B$ relatively easy. – Daniel Fischer Nov 14 '13 at 15:49
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    $A^B=I$ implies that $B=A^{-}$ ($A$ is necessarily non-singular), so that the first row $b_1^$ of $B$ is $b_1^=e_1^TB=e_1^TA^{-*}$, i.e., $b_1$ is the solution of $Ab_1=e_1$ ($e_1$ is the first vector of the standard basis). – Algebraic Pavel Nov 14 '13 at 20:27

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