Consider a system of linear equations $$A \times x = B$$
The system has a unique solution exactly when the determinant of the coefficient matrix (i.e. A) is nonzero. When the determinant of the coefficient matrix is zero, there are either no solutions or many solutions.
For overdetermined system of linear equations, If no. of independent equations is equal to the no. of unknowns, then it has a unique solution.
Now consider the example
$$2x_1 + x_2 = 3$$ $$2x_1 - x_2 = 5$$ $$x_1 - 2x_2 = 4$$
Above system of equations can be written as
$ \begin{bmatrix} 2 & 1 & 0 \\ 2 & -1 & 0 \\ 2 & -2 & 0 \end{bmatrix} \times $$ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} $ $=\begin{bmatrix} 3 \\ 5 \\ 4 \end{bmatrix}$
As we can see from the above graph, the system of linear equation has a unique solution. But at the same time the
$ \begin{vmatrix} 2 & 1 & 0 \\\ 2 & -1 & 0 \\\ 2 & -2 & 0 \end{vmatrix} = 0 $
Not even this one, determinant of all overdetermined system of linear of equation will have determinant of coefficient martix as zero, because last, 2nd last, .. column vectors will have all zero's. Why above two definition are contradicting each other, What's the thing that I'm missing here.
It's not unique, because $x_3$ can be anything.