linear algebra:prove or disprove Does there exist a matrix of order $n\times n$ above $F$, say $A$, so that for every $\underline{b}\in F^n$ there are infinite solutions to A$\underline{x}$ = $\underline{b}$?
 A: No, because if such matrix exist it will be surjective (by definition a surjective map is a maps who for all $B$ it exist a least one solution $X$ of $AX=B$) and because your matrix is square the matrix will by injective that mean $AX=0$ have only $0$ as solution.
Contradiction 
A: The answer is no. Suppose by contradiction that such a matrix $A$ exist. by Our assumption, the system $Ax=b$ has a solution for every $b$. In particular it has a solution $x_i$ for
$$
b_i=
\begin{pmatrix}
0 \\ \vdots \\ 1 \\ \vdots \\ 0 
\end{pmatrix},
$$
where $b_i$ has $1$ in its $i$-th position and zeros elsewhere. Let $X$ be the matrix that $x_i$ is in its $i$-th column: 
$$
X=\begin{pmatrix}
\mid & \cdots & \mid \\
x_1  & \cdots & x_n\\
\downarrow & \cdots & \downarrow \\
\end{pmatrix},
$$
Since $Ax_i=b_i$, it follows that 
$$
AX=\begin{pmatrix}
\mid & \cdots & \mid \\
b_1  & \cdots & b_n\\
\downarrow & \cdots & \downarrow \\
\end{pmatrix}, 
$$
that is $AX=I_n$. Thus, $A$ is invertible. But then, the system $AX=b$ has a unique solution for every $b$, a contradiction. 
A: The answer is $\color{red}{no}$. If there is such a matrix and we consider the columns of $A$ as the vector $b$, you see that there are infinite inverses for matrix $A$ while we know the inverse of each matrix is unique. So, there is no such a matrix.
