# Is it possible for a triangular matrix in echelon form to not have a unique solution and how?

I want to know if it is possible for a triangular matrix in echelon form to not have a unique solution and how?

Isn't there something to do with the determinant that shows this? or am I wrong?

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Note that a triangular matrix may have zeros on the diagonal, for example, $$\pmatrix{1&2&3\cr 0&4&5\cr0&0&0\cr}\ .$$ A system with this echelon form for its left hand side will have zero determinant, and will have infinitely many solutions or no solution, depending on the right hand side.

For an extreme example, the matrix which has every entry zero also counts as a triangular matrix.

However, if you have a system in which the left hand side is represented by a square matrix, and if you reduce it to echelon form (which will be triangular), and if it then does not have any zeros on the diagonal, then there will be a unique solution.

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