Prove or Disprove the Existence of Solutions...

Question

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)

1. $Ax=b$ has an integer solution.
2. $Ax=b$ has a rational solution.
3. The set of real solutions to $Ax=0$ has a basis consisting of rational solutions.
4. 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 ?

Answer 1

$1.$ 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.

Answer 2

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.