0
$\begingroup$

Suppose I have a system of equations whose augmented matrix is $$ \left[\begin{array}{ccc|c} 1 & 2 & 3 & 7\\ 4 & 5 & 6 & 8 \end{array}\right] $$ After converting this matrix to row reduced echelon form I got $$ \left[\begin{array}{ccc|c} 1 & 0 & -1 & -19/3 \\ 0 & 1 & 2 & 20/3 \end{array}\right] $$

Now I want a solution such that $$0 < x_1 < 9, \quad 0 < x_2 < 7, \quad 0 < x_3 < 2$$ One such solutions is $x_1 = -16/3 , x_2 = 14/3$ and $x_3 = 1$.

But how do I find one such solution in general case, suppose I have n unknowns variables ? I am only considering the case when infinite solutions exist for a linear system

$\endgroup$
1
  • $\begingroup$ What logic should I use in coding to get one desired solution. $\endgroup$ Sep 16, 2018 at 12:28

1 Answer 1

0
$\begingroup$

You determined that the null space of your system is $\operatorname{span}\left\{\begin{bmatrix} 1 \\ -2 \\ 1\end{bmatrix}\right\}$ so the general solution is

$$x = \begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix} -\frac{19}3 \\ \frac{20}3 \\ 0\end{bmatrix} + t\begin{bmatrix} 1 \\ -2 \\ 1\end{bmatrix} = \begin{bmatrix} -\frac{19}3 + t \\ \frac{20}3-2t \\ t\end{bmatrix}$$

Therefore you can solve for $t$ independently in $$0 < -\frac{19}3 + t < 9 \implies t \in\left\langle \frac{19}3, \frac{100}3\right\rangle$$ $$0 < \frac{20}3-2t < 7 \implies t \in\left\langle -\frac16, \frac{10}3\right\rangle$$ $$0 < t < 2 \implies t \in \langle 0, 2\rangle$$

an take the intersection.

$\endgroup$
2
  • $\begingroup$ I am asking for general case where number of free variables are more, not just one. I can't use nested loops for every variable. $\endgroup$ Sep 16, 2018 at 12:57
  • $\begingroup$ How should I code this if I have more than one free variable. I dont't want to use many nested loops. $\endgroup$ Sep 16, 2018 at 13:22

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.