# Solutions of a linear system in RREF

So I was given a reduced row echelon matrix that corresponds to this system:

$$0x_1 + x_2 + 0x_3 = 7$$

$$0x_1 + 0x_2 + x_3 = 2$$

That is, the first two elements of each row were 0.

The question was, determine the leading and free variables and find the solution set.

So, technically, $$x_1$$ is a free variable, but the system essentially is

$$x_2 + 0x_3 = 7$$

$$0x_2 + x_3 = 2$$

Then, can I still call $$x_1$$ a free variable? And, if so, the solution set will be ($$x_1$$, 7, 2), for all $$x_1$$ in R, or just (7, 2)?

The augmented matrix describing our system is $$\left[\begin{array}{rrr|r} 0 & 1 & 0 & 7 \\ 0 & 0 & 1 & 2 \end{array}\right]$$ The pivots in the coefficient matrix correspond to the variables $$x_2$$ and $$x_3$$. This means that $$x_2$$ and $$x_3$$ are the "dependent" variables and $$x_1$$ is the "free" variable.
The solutions to the system are given by $$\left[\begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] = \left[\begin{array}{r} x_{1} \\ 7 \\ 2 \end{array}\right] = x_1\left[\begin{array}{r} 1 \\ 0 \\ 0 \end{array}\right] +\left[\begin{array}{r} 0 \\ 7 \\ 2 \end{array}\right]$$
The pivot columns are the columns corresponding to $$x_2$$ and $$x_3$$, and thus these two variables are so-called basic variables. Variables that are not basic are called free variables. In this case, as $$x_1$$ doesn't correspond to a pivot column, it is a free variable. The solution of this system of linear equations is specified by three variables, so we would write $$(x_1,7,2)$$ for any $$x_1\in \mathbb{R}$$.