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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)?

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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 $$ \left[\begin{array}{r} x_{1} \\ x_{2} \\ x_{3} \end{array}\right] = \left[\begin{array}{r} x_{1} \\ 7 \\ 2 \end{array}\right] = x_1\left[\begin{array}{r} 1 \\ 0 \\ 0 \end{array}\right] +\left[\begin{array}{r} 0 \\ 7 \\ 2 \end{array}\right] $$

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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 $x_1\in \mathbb{R}$.

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