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I would like to solve the inequality, $$Ax≥b⟺Ax=b+z,z≥0⟺x=A^{−1}b+A^{−1}z,z≥0$$

$$ z_i = (Ax)_i - b_i, \, \forall i $$, here $A$ is a matrix and $x$ and $b$ are vectors.

Here is a minimal problem that I would like to solve -

$\begin{pmatrix} 2 & 0 \\ 0 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \geq \begin{pmatrix} 4 \\ 6 \end{pmatrix}$

Please let me know if something is unclear in my question.

Regarding my prerequisite knowledge, I have a Master's in Engineering. During that time, I did touch on some linear programming concepts like Simplex. Unfortunately, I don't remember much about it.

desert_ranger
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    What does it mean to be greater than $\begin{pmatrix} 4 \ 6 \end{pmatrix}$? – Zubin Mukerjee May 22 '24 at 14:34
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    @ZubinMukerjee It means that $x \geq 4$ and $y \geq 6$. Please let me know if you need more clarification. – desert_ranger May 22 '24 at 14:37
  • You have no inequalities between vectors, you could only compare there length. if you compare the components it just gives you for example 2x>4 This does not seem a problem? – trula May 22 '24 at 14:40
  • I did not know that there are no inequalities between vectors. Thank you for the information.

    Yes, I understand that comparing components will lead to scalar equations. Would there be a problem doing that? Can't these scalar equations once again be placed back in Matrix form?

     – desert_ranger May 22 '24 at 14:43
  • Componentwise inequality as you've described does make sense, and is usually denoted by $\prec$ and $\preceq$. See this answer for discussion of the notation: https://math.stackexchange.com/a/669089 – Zubin Mukerjee May 22 '24 at 14:47
  • @ZubinMukerjee Thank you for your answer. However, I find that with your answer, I won't be able to combine my results, back into a matrix. I asked a similar question here: https://math.stackexchange.com/questions/4919278/what-algorithm-can-solve-linear-inequalities-when-dealing-with-matrices and was asked to use linear programming. – desert_ranger May 22 '24 at 14:49
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    Why do you want to combine your results back into a matrix? What is the original problem you are working on? What is the goal? You can combine them back into a componentwise inequality if you want - this is only notation, as it means the same thing as: $x \geq 2$ and $-y \geq 2$. I've added that to the answer. Remember to flip inequalities when multiplying/dividing by negative constants – Zubin Mukerjee May 22 '24 at 15:00
  • @ZubinMukerjee Thank you for your edit. The reason I want to combine the results back into a matrix is because computing $\begin{pmatrix} x \ y \end{pmatrix}$ is only a part of my algorithm. I need to process it further. Additionally, I also need an algorithmic way to perform the computation shown in your answer. – desert_ranger May 22 '24 at 16:23

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Assuming you are referring to componentwise inequality, usually denoted by $\preceq$, you can multiply:

$$\begin{align} \begin{pmatrix} 2 & 0 \\ 0 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} &\succeq \begin{pmatrix} 4 \\ 6 \end{pmatrix}\\\\ \begin{pmatrix} 2x \\ -3y \end{pmatrix} &\succeq \begin{pmatrix} 4 \\ 6 \end{pmatrix} \end{align}$$

So,

$$2x \geq 4 \qquad \implies \quad x \geq 2$$ $$-3y \geq 6 \qquad \implies \quad y \leq -2\\$$

Notice that the inequality in the last line is flipped!

If you prefer, you can write the final inequalities as:

$$\begin{pmatrix} x \\ -y \end{pmatrix} \succeq \begin{pmatrix} 2 \\ 2 \end{pmatrix}$$

Zubin Mukerjee
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