System of linear equations: When are we guaranteed a unique solution?

[kent davidge]

I was reading about systems of linear equations. Even if we have the same number of unknowns and equations, we may still have infinitely many or no solutions. But if in addition to that the determinant of the matrix of coefficients does not vanish, then does it necessarily imply that we have a unique solution?

Surprisingly or not, I haven't found an answer to the above question.

 

[fresh_42]

Yes, since we then can divide: $\mathbf{A}\mathbf{x}=\mathbf{b} \Longrightarrow \mathbf{x}=\mathbf{A^{-1}}\mathbf{b}.$

https://en.wikipedia.org/wiki/System_of_linear_equations#Matrix_solution
https://en.wikipedia.org/wiki/Invertible_matrix#In_relation_to_its_adjugate

 

[Leo Liu]

https://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-b-matrices-and-systems-of-equations/session-14-solutions-to-square-systems/MIT18_02SC_MNotes_m3.pdf

 

[nuuskur]

For the system $Ax = b$ with $A\in\mathrm{Mat}_{m, n}(K)$, there are $n-\mathrm{rank}(A)$ free variables. If $K=\mathbb R$ (or any other infinite field), then having at least one variable free means there are infinitely many solutions. Also, a solution can be unique only if the matrix $A$ is a square matrix (i.e only if number of variables = number of equations).

Of course, if the underlying field is finite, there can't be infinitely many solutions. That's something a lot of my students get wrong. They remember from school that if number of eqs is smaller than number of variables, then there are automatically infinitely many solutions. Not necessarily.

 

[mathwonk]

This is a nice answer, but I think you are thinking of the case b = the zero vector. I.e. probably you need to assume also the system is consistent. As you stated it, it seems, even with an infinite field and some free variables, if b ≠ zero, you only get either infinitely many, or no solutions. E.g. you could have the system x+y = 0, and x+y = 1, where the rank of the 2x2 matrix A is one, hence there is one free variable, but there is no solution at all, since the vector (0,1) does not lie in the image {(t,t): all t in K}, of the linear transformation.

 

[nuuskur]

Good point. The system $Ax=b$ has a solution if and only if $\mathrm{rank}(A) = \mathrm{rank}(A|b)$ by Kronecker-Capelli.