Prove or Disprove the Existence of Solutions... 
Let $A$ be a $3\times 4$ and $b$ be a $3\times 1$ matrix with integer entries.Suppose that the system $Ax=b$ has a complex solution. Then which of the following are true? (CSIR December 2014)
  
  
*
  
*$Ax=b$ has an integer solution. 
  
*$Ax=b$ has a rational solution.
  
*The set of real solutions to $Ax=0$ has a basis consisting of rational solutions.
  
*If $b$ is not equal to zero then $A$ has positive rank.
  

Is it possible to say that 1 and 2 are true as $Ax=b$ has a complex solution? If not  what we have to understand from the statement "the system $Ax=b$ has a complex solution " ? What about 3 and 4  ? 
 A: $1.$ is false. consider the following counter example.
\begin{align}
\begin{pmatrix}
2 & 1 & 0 & 0\\
 0& 2 & 6 & 0\\
0 & 0 & 1 & 0
\end{pmatrix}\begin{pmatrix}
              x \\ y \\ z \\ w
             \end{pmatrix} = \begin{pmatrix}
              1 \\ 3 \\ 0 
             \end{pmatrix} 
\end{align}
$2.$ is always true. Note that as $Ax=b$ has a complex solution, so $A$ and the augmented matrix $A|b$ has same column rank. Now using row reduced echelon form, whose entry in this case will be rational number, one will always have a rational solution.
$3.$ is also true using the row reduced echelon form of the augmented matrix $A|b$ one can construct a basis whose entry are rational.
$4.$ is obviously true. 
A: Let $A$ be a $3×4$ and $b$ be a $3×1$ matrix with integer entries.Suppose that the system $Ax=b$ has a complex solution. Let this complex solution is $u$= 
$
  \left( \begin{array}{ccc}
x_1+i\times y_1  \\
x_2+i\times y_2 \\
x_3+i\times y_3\end{array} \right) $ 
=
$
  \left( \begin{array}{ccc}
x_1  \\
x_2 \\
x_3\end{array} \right) $ + 
$i
  \left( \begin{array}{ccc}
y_1  \\
y_2 \\
y_3\end{array} \right) $= $X+i \times Y$ ( where $Y$ is not a zero vector)
So this system of linear equation can be written as: $A(X+i \times Y)=b +0$ since entries of $b$ are integers..i.e., real so we can consider it as two system of linear equations: $A X=b$ and $AY=0$   that means $AX=0$ has a solution other than zero. (which is $Y$). So it is clear that null space has dimension greater than equals to 1. Because $Y $ belong to Null Space. So 3rd option is correct.  1st option is not correct. you can verify by taking any particular example.
