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Take some matrix $A$ such that

$$ A = \begin{bmatrix} \alpha_{11} & \alpha_{12} & \dots & \alpha_{1n} \\ \alpha_{21} & \alpha_{22} & \dots & \alpha_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ \alpha_{n1} & \alpha_{n2} & \dots & \alpha_{nn} \end{bmatrix}, $$

where each $\alpha_{ij} \in \mathbb{Z}$, and assume $\det(A)$ is nonzero. Also take column vectors $\mathbf{x} = (x_1, \dots, x_n)^T, \mathbf{b}= (\beta_1, \dots, \beta_n)^T$, where $\beta_i \in \mathbb{Z}$ and the $x_i$'s are unknown. Consider the matrix equation

$$ A\mathbf{x} = \mathbf{b} $$

which is equivalent to the system of linear equations

$$ \sum_{j=1}^{n} \alpha_{ij} x_j = \beta_i, $$

for $1 \leq i \leq n$. I understand that the equation can be solved via Gaussian reduction of $A$, however, I was wondering if there was an explicit way to calculate each $x_i$ in terms of the $\alpha_i, \beta_i$ by repeated substitution?

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  • $\begingroup$ Yes, you can substitute, but it will be considerably more work. The most efficient known exact method of solving is still Gaussian elimination with back substitution. $\endgroup$ May 14, 2018 at 17:54
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    $\begingroup$ What you're wanting to do looks like it will boil down to Cramer's rule. $\endgroup$
    – Ivo Terek
    May 14, 2018 at 17:57
  • $\begingroup$ @IvoTerek That's what I was looking for - +1! $\endgroup$
    – Chris
    May 14, 2018 at 18:14
  • $\begingroup$ Ok, then I'll convert the comment to an answer and you can accept it (so the question does not stay in the unanswered list) :-) $\endgroup$
    – Ivo Terek
    May 14, 2018 at 18:18

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The procedure you're looking for is Cramer's rule.

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    $\begingroup$ thanks a lot mate! $\endgroup$
    – Chris
    May 14, 2018 at 18:34

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