How to frame this set of linear equations?

Question

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?

Answer 1

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.