Question regarding determinants and switching columns of square matrices

$\displaystyle A$ is a square matrix length n and $\displaystyle b$ is a vector in $\displaystyle R^n$. For every $\displaystyle k$=1,2,...,n $\displaystyle A_k$ is the matrix that is received from $\displaystyle A$ after switching the $\displaystyle k$ column with vector $\displaystyle b$. Given, detA=0.

Prove that if det$\displaystyle A_1 \neq 0$ then there is no solution to $\displaystyle Ax=b$.

This is what the question means if it was unclear:

A = $\displaystyle \left(\begin{array}{cccc}\alpha_{11} & \alpha_{12} & ... & \alpha_{1n}\\ \alpha_{21} & \alpha_{22} & ... & \alpha_{2n}\\ . & . & . & . \\ \alpha_{n1} & \alpha_{n2} & ... & \alpha_{nn} \end{array}\right)$

b = $\displaystyle \left(\begin{array}{cccc} \beta_1 \\ \beta_2 \\ . \\ \beta_n \end{array}\right)$

$\displaystyle A_1 = \left(\begin{array}{cccc}\beta_1 & \alpha_{12} & ... & \alpha_{1n}\\ \beta_2 & \alpha_{22} & ... & \alpha_{2n}\\ . & . & . & . \\ \beta_n & \alpha_{n2} & ... & \alpha_{nn} \end{array}\right) \rightarrow $ I simply replace the first column of A with vector b

Now that it's all cleared up how do I solve this?? It has something to do with detA=0 which makes the matrix singular... and most likely it has to do something with all those properties of an invertibale matrix.

And the question is telling us that there is only a solution to Ax=b if A_1 is singular like A! Why is that so? Why can't A_1 be invertible?

But I just can't put it together. Can someone please help?

Thanks!