Chapter 1 - Linear Equations in Linear Algebra - 1.3 Exercises - Page 33: 22

$$ A = \begin{bmatrix} 1 & 2 & 3 \\ 1 & 2 & 3 \\ 1 & 2 & 3\\ \end{bmatrix} $$ $$b=\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$$

$A$ is $3 \times 3$ and has no entry whose value is $0$. $b$ is clearly not in the column space of $A$ because it is not a scalar multiple of the first column (the second and third columns are multiples of the first column).