how det(A)=0 implies that the solution isn't unique? Solution of matrix equation Ax=b, where
$$ A=\left(\begin{matrix}
a_1&a_2&\dots&a_n
\end{matrix}\right), \ a_i \in \mathbb{R}^n,$$
is not unique, if vectors $$ a_1, \ a_2, \dots, \ a_n $$
are linearly dependent. Then by properties of determinant, $$ \det A=0. $$
However, does it always follow, that if det A = 0, column vectors of A are linearly dependent? Can someone present a proof?
 A: One possible proof:

*

*Suppose the columns are linearly independent.

*Convert the matrix into a column echelon form, starting from the last column and working your way backwards.

*You know the number of linearly independent columns is the number of nonzero columns you end up with. However, as you've assumed the columns are independent, there are no zero columns.

*In other words, you've ended up with a triangular matrix with all nonzero elements on the diagonal. Its determinant is nonzero.

*However, the elementary transformations we use when converting the matrix into a row/column echelon form don't change the property of the diagonal to be zero or nonzero.

*Thus, the determinant was nonzero to start with.

A: If the first column is all $0$'s, clear. Otherwise, consider a row with first element $\ne 0$. Permute it so it becomes the first row. The determinant is still $0$, the  system is equivalent to the previous. Now reduce all elements in first column, lower than the first row. Determinant still $0$, system still equivalent. Now, look at the matrix formed by removing first row and column. Determinant is $0$. Apply induction, find a non-zero solution $(x_2, \ldots, x_n)$. Now use the original first equation to get $x_1$. Now we have a non-zero solution to the whole system.
