I know that to solve a linear equation involving $n$ variables for example, we need $n-1$ other independent equations to form a system and then solve that system. Is there a formal proof for this?
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$\begingroup$ It depends on the type of equation. Are they all linear? Occurring over a field? $\endgroup$– RandallFeb 21, 2019 at 18:32
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$\begingroup$ I'm talking about linear equations $\endgroup$– GDGDJKJFeb 21, 2019 at 18:33
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$\begingroup$ The solution can only be unique , if we have at least as many equations than variables. This can be shown by looking at the poosible ranks of the coefficient matrix. $\endgroup$– PeterFeb 21, 2019 at 18:44
2 Answers
Proof: if we didn't have n equations for n unknowns, then the matrix representing the system would either be rectangular (meaning no unique solution) or have zero rows (also no unique solutions). Therefore there must be n equations for n unknowns in order for a unique solution to exist.
If it's a system of linear equations, then you can use Gauss-Jordan elimination to get the solutions. You can also take the determinant of the coefficient matrix first to check whether or not you'll have a single solution, infinitely many solutions, or zero solutions.