# 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)

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 ?

• 4) is true since $b$ is in the image of $A$, and if $b\ne 0$, then the image of $A$ is nontrivial, so the rank is positive. – jgon May 7 '15 at 6:25
• 1) is definitely false, since it includes as a special case the equation $2x=1$. As for 2) and 3): think about what happens to the entries of the augmented matrix when you perform elementary row operations.... – Greg Martin May 7 '15 at 6:42
• is it be fixed or can be variable with integer entries? – Piquito May 7 '15 at 15:25
• I think that $b$ is $4 \times 1$ instead of $3 \times 1$... – Ritu May 23 '15 at 7:29
• No it is $3\times 1$... – Tani May 23 '15 at 7:33

$1.$ 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.

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.