11
$\begingroup$

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 ?

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – jgon May 7 '15 at 6:25
  • 2
    $\begingroup$ 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.... $\endgroup$ – Greg Martin May 7 '15 at 6:42
  • $\begingroup$ is it be fixed or can be variable with integer entries? $\endgroup$ – Piquito May 7 '15 at 15:25
  • $\begingroup$ I think that $b$ is $4 \times 1$ instead of $3 \times 1$... $\endgroup$ – Ritu May 23 '15 at 7:29
  • $\begingroup$ No it is $3\times 1$... $\endgroup$ – Tani May 23 '15 at 7:33
7
+50
$\begingroup$

$1.$ 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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.