Set notation feasible region to matrix vector form

[TheFallen018]

Hey, I'm a little unsure about this question. I've got an answer, but I'm not sure that it's correct, and I can't find the info I need relating to this.

Here is the question

Write the feasible region for
$\displaystyle C \subset R^{3}$, where
$\displaystyle C=((x,y,z): x,y,z\geq 0,5x+z\leq 8,z-2y\leq13,x-y-z\geq-3))$,
in matrix-vector form
$\displaystyle \mathbf{Ax \leq b}$.

What I thought, is just make each equation in the set have the same inequality, and then put them in a matrix and a vector, as so

\(\displaystyle



\begin{pmatrix}
-1 &0 &0 \\
0 &-1 &0 \\
0 &0 &-1 \\
5 &0 &1 \\
0 &-2 &1 \\
-1 &1 &1
\end{pmatrix}\leq\begin{pmatrix}
0\\
0\\
0\\
8\\
13\\
-3\\
\end{pmatrix}\)


But this feels very wrong. Could anyone point me in the right direction?

Thanks

 

[HallsofIvy]

Obviously it makes no sense to say that a 3 by 6 matrix is "less than or equal to" 6-vector. Actually, I would say that it makes no sense to say that one vector is "less than or equal to" another but here they are clearly using it "component wise" (which would not work for a general vector space).

You are asked to write the system of inequalities in the form $\displaystyle Ax\le b$ but you forgot the "x"! You just have $\displaystyle A\le b$.

 

[TheFallen018]

Obviously it makes no sense to say that a 3 by 6 matrix is "less than or equal to" 6-vector. Actually, I would say that it makes no sense to say that one vector is "less than or equal to" another but here they are clearly using it "component wise" (which would not work for a general vector space).

You are asked to write the system of inequalities in the form $\displaystyle Ax\le b$ but you forgot the "x"! You just have $\displaystyle A\le b$.

So, do I just need to put x back in there?

\(\displaystyle



\begin{pmatrix}
-1 &0 &0 \\
0 &-1 &0 \\
0 &0 &-1 \\
5 &0 &1 \\
0 &-2 &1 \\
-1 &1 &1
\end{pmatrix}\begin{pmatrix}
x\\
y\\
z\\
x\\
y\\
z\\
\end{pmatrix}\leq\begin{pmatrix}
0\\
0\\
0\\
8\\
13\\
-3\\
\end{pmatrix}\)

However, this doesn't seem right since A isn't linearly independent. Can I drop some of the inequalities so that A is 3x3?