Gaussian Elimination with Vectorial Ineqaulity

Question

I have a inequality relation;

$$ \begin{bmatrix} -0.1619 && 0.3417 && -0.1799 \\ 0 && 0 && 0 \\ 0 && 0 && 0 \\ \end{bmatrix} \vec{R} \succeq \vec{0} $$

Where $\succeq$ is vectorial inequality (all members of the vector are greater than or equal to the right hand side). This gives me the non-trivial relation

$$-0.1619R_1 + 0.3417R_2 - 0.1799R_3 >= 0\tag{1}$$

If instead, I perform Gaussian elimination to get the matrix to row-echelon form, I get

$$ \begin{bmatrix} 1 && -2.111 && 1.111 \\ 0 && 0 && 0 \\ 0 && 0 && 0 \\ \end{bmatrix} \vec{R} \succeq \vec{0} $$

which gives the relation

$$R_1 + -2.111R_2 +1.111R_3 >= 0\tag{2}$$

I know from checking with a given solution $R = \left[0 \; 2 \; 1\right]^T$ that the (1) is the correct solution, but I don't understand why (2) isn't right. Can you not perform Gaussian elimination row operations with inequality signs?

Answer 1

You have multiplied by a negative number , $-\frac1{0.1619}$, hence the inequality is flipped.

Also let's consider another example:

$x_1 \ge 0, x_2 \ge 0$

suppose you replace it by

$x_1+x_2 \ge 0, x_2 \ge 0$

We can see that we are describing a different polyhedron though the first set of constraints does imply the second set of constraints.