# System of linear simultaneous equations

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?

• 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. May 14, 2018 at 17:54
• What you're wanting to do looks like it will boil down to Cramer's rule. May 14, 2018 at 17:57
• @IvoTerek That's what I was looking for - +1! May 14, 2018 at 18:14
• 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) :-) May 14, 2018 at 18:18