How to frame this set of linear equations? I have the following set of equations, as an example
$2x + 1y + 2z = A$
$0x + 2y + 2z = A$
$1x + 2y + 1z = A$
I assume this can be rewritten as a matrix? How can I check if a solution exists such that x, y, and z are nonnegative? In this case I don't believe a solution exists but how can I verify it without manually testing values?
 A: Use an augmented coefficient matrix, and obtain row-echelon form (using elementary row operations), to see if a solution exists, and/or if the system is inconsistent. If inconsistent, then no solution exists.
$2x + 1y + 2z = A$
$0x + 2y + 2z = A$
$1x + 2y + 1z = A$
$$
M = \begin{pmatrix}
2 & 1 & 2 & A \\ 
0 & 2 & 2 & A \\
1 & 2 & 1 & A
\end{pmatrix}
$$


*

*Subtract 1/2 × (row 1) from row 3

*Multiply row 3 by 2

*Swap row 2 with row 3

*Subtract 2/3 × (row 2) from row 3

*Multiply row 3 by 3

*Subtract 1/3 × (row 3) from row 1

*Subtract 1/3 × (row 2) from row 1

*Divide row 1 by 2

*Divide row 2 by 3

*Divide row 3 by 6


$$\text{Result}:\quad
\begin{pmatrix}
1 & 0 & 0 & A/6\\
0 & 1 & 0 & A/3\\
0 & 0 & 1 & A/6
\end{pmatrix}
$$
If you row reduce carefully, (and you should attempt this so you can gauge your success in being able to do so), you should obtain the following:
$$x = A/6,\; y = A/3, \; z = A/6$$
So for any given value of A, you will have a unique solution for $\begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} A/6 \\ A/3 \\ A/6 \end{pmatrix}$.
So long as $A\geq 0$, the solution will be non-negative.

For review: see row echelon form and row operations.
