Linear Algebra solution when determinant is zero I am doing practice questions in my book and I came upon this True/False question: 
If $\det(A) = 0$, then the linear system $Ax=b$, $b\neq 0$, has no solution. 
The book is saying that the answer is false. But why is that? I thought the answer is true because of something like this 
$  Ax  = b$ 
$$\left(\begin{array}{ccc|c}1&0&0&1\\0&0&0&2\\0&0&1&3\end{array}\right)$$
When a matrix has its rref taken, the resulting matrix, when the determinant is zero, would always have a zero in its diagonal, right? This would result in a matrix with no solution because row 2 is impossible. Am I misunderstanding the question somehow? I am also confused by this question because I am not sure how augmented matrices work with square matrices because you can only find the determinant of a square matrix. Can someone please explain why the answer is false?
 A: The statement is not true. For it to be true, it would have to be true for any combination of $A$ and $b$ where $\det(A)=0$ and $b\neq 0$, but that is not the case, since the equation 
$$\begin{bmatrix}1 & 0\\0&0\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}1\\0\end{bmatrix}$$
clearly has a solution
A: In your example, if you choose $b$ equal to one of the colums of $A$, then the system $Ax=b$ is solvable.
If $\det(A)=0$ then the system $Ax=b$ is never uniquely solvable, though. So you have the alternatives: no solution or many solutions (depends on the underlying field).
A: Excluding the already mentioned cases where we had one line filled with zeros in the complete matrix, we don't need to have a zero line in order to have $\textrm{det}(A)=0$.
Just take the cases where one line is a multiple from the other, for example:
$$ \begin{bmatrix}a & b\\ka&kb\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}c\\kc\end{bmatrix} $$
This is a system which has $\textrm{det}(A)=0$ and also has infinite solutions as long $a$ and $b$ are not both zeros (except in the case $c$ is also zero).
A: If $\det A=0$, then there are indeed vectors $b$ for which $Ax=b$ does not have a solution.
But, there are also values of $b$ for which it DOES have a solution. Say $a$ is an arbitrary vector and $b=Aa$. Then clearly the system $Aa=b$ possesses at least one solution, namely $x=a$. 
