Let the system $AX=B$, where $$A=\begin{bmatrix}8&3\\2&6\end{bmatrix} \ \text{and} \ B=\begin{bmatrix}3\\-1\end{bmatrix}.$$
The book says it does not have a solution in $\mathbb{F}_{2}$ and $\mathbb{F}_{3}$ but have a solution in $\mathbb{F}_{7}$, even though $\det A≡ 0 \pmod{7}$. Can someone please explain how it's possible?
As in all fields, a linear system $AX=B$, where $A$ is a square matrix, has solutions in the field $F$ if and only if the rank of $A$ is equal to the rank of the augmented matrix $[A | B\,]$. Here is how the augmented matrix can be rewritten in each field: $$\begin{matrix} \mathbf F_2 & \mathbf F_3 & \mathbf F_7 \\ \left[\begin{array}{cc|c} \mkern-4mu 0&1& 1\mkern-6mu\\\mkern-4mu 0&0&1\mkern-4mu \end{array}\right] &\quad \left[\begin{array}{cc|c} \mkern-4mu 2 & 0 & 0\mkern-6mu\\\mkern-4mu 2&0& 2\mkern-4mu \end{array}\right]\quad & \left[\begin{array}{cc|c} \mkern-4mu 1 & 3 & 3\mkern-6mu\\\mkern-4mu 2&6&6\mkern-4mu \end{array}\right] \end{matrix}$$
