# How to write the solution set of an augmented matrix in reduced-row echelon form?

I am new here, so please excuse me if my question is not on par with the guidelines (which I have read).

Introduction

I am used to having one variable in the tuple, e.g.:

$$\begin{array}{rcrcrcr} x & & & + & \frac{3}{5}z & = & 1\\ & & y & - & \frac{2}{5}z & = & -1\\ & & & & 0 & = & 0 \end{array}$$

Would have the solution set as a set of tuples:

$$S=\left\{ \left(1-\frac{3}{5}z,-1+\frac{2}{5}z,z\right)\mid z\in\mathbb{R}\right\}$$

And a solution set as a set of vectors:

$$S=\left\{ \begin{pmatrix}1\\ -1\\ 0 \end{pmatrix}+\begin{pmatrix}-\frac{3}{5}\\ \frac{2}{5}\\ 1 \end{pmatrix}z\mid z\in\mathbb{R}\right\}$$

My Question

$$\begin{array}{rcrcrcr} x & + & y & - & z & = & 3\\ 2x & & & - & z & = & 1\\ 3x & + & y & + & z & = & 0 \end{array}$$

Write it as an augmented matrix and perform row operations to produce a version in reduced-row echelon form:

$$\left[\begin{array}{rcrcrc|r} 1 & 0 & 0 & -\frac{1}{6}\\ 0 & 1 & 0 & \frac{11}{6}\\ 0 & 0 & 1 & -\frac{4}{3} \end{array}\right]$$

The solution set's 3-tuple is:

$$\left(-\frac{1}{6},\frac{11}{6},-\frac{4}{3}\right)$$

1. How do I write the solution set as a set of tuples?
2. How do I write the solution set as a set of vectors?

Thank you!

The solution set can be written as a set of tuples like this:

$$\Big\{\big(-1/6, 11/6, -4/3\big)\Big\}$$

Or as a set of vectors like this:

$$\Bigg\{\begin{bmatrix}-1/6 \\ 11/6 \\ -4/3\end{bmatrix}\Bigg\}$$

In general, when there is only one solution, then the solution set is simply a set with one element. Easy peasy!

• Got it. I wasn't sure because I am used to writing $\mid\rm{variable(s)}\in\mathbb{R}$. Commented Sep 27, 2020 at 19:38
• @Oliver Yep, it sounds like you’re most familiar with set builder notation, but that’s only one way of specifying sets. The other most common way is to simply list out its elements between curly braces, if it has finitely many. Commented Sep 27, 2020 at 19:41
• UPDATE: This comment is dumb, because there is only one solution. So, if I were to write it in set builder notation, would there be a $\mid\rm{variable(s)}\in\mathbb{R}$? Commented Sep 27, 2020 at 19:42
• @Oliver Lol, that’s kind of a funny question, but sure. You could alternatively write the solution as $$\{(-1/6, 11/6, -4/3)|x\in\mathbb R\}$$. It would be equal to the set specified in my answer, though - the free variable $x$ has no bearing on what is to the left of the |, so it’s extraneous and unnecessary. You could also write $$\{(-1/6, 11/6, -4/3)|x,y,z\in\mathbb R,\space a,b\in\mathbb Q, \space p,q,r\in\mathbb N\}$$ and it would be exactly the same. Commented Sep 27, 2020 at 19:49
• Thank you for all your help. I wish I knew enough about the subject to understand why my question is funny! Or perhaps it was the fact that I was trying to make it more complicated than it needs to be? Commented Sep 27, 2020 at 20:09

You seem to have a problem with there being only one solution rather than infinitely many solutions. Most people would consider that a good thing!

The "solution set" is the singleton set $$\{\left(-\frac{1}{6}, \frac{11}{6}, -\frac{4}{3}\right)\}$$. The "set of vectors" is again a singleton set,$$\{-\frac{1}{6}\vec{i}+ \frac{11}{6}\vec{j}- \frac{4}{3}\vec{k}\}$$.