##### Example1.2.5Describing the solution space from a reduced row echelon matrix

1. Consider the row reduced echelon matrix

\begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 2 \amp -1 \\ 0 \amp 1 \amp 1 \amp 2 \\ \end{array} \right]. \end{equation*}

Its corresponding linear system may be written as

\begin{equation*} \begin{alignedat}{4} x \amp \amp \amp {}+{} \amp 2z \amp {}={} \amp -1 \\ \amp \amp y \amp {}+{} \amp z \amp {}={} \amp 2. \\ \end{alignedat} \end{equation*}

Let's rewrite the equations as

\begin{equation*} \begin{alignedat}{2} x \amp {}={} \amp -1 -2z\\ y \amp {}={} \amp 2-z. \\ \end{alignedat} \end{equation*}

From this description, it is clear that we obtain a solution for any value of the variable $$z\text{.}$$ For instance, if $$z=2\text{,}$$ then $$x = -5$$ and $$y=0$$ so that $$(x,y,z) = (-5,0,2)$$ is a solution. Similarly, if $$z=0\text{,}$$ we see that $$(x,y,z) = (-1,2,0)$$ is also a solution.

Because there is no restriction on the value of $$z\text{,}$$ we call it a free variable, and note that the linear system has infinitely many solutions. The variables $$x$$ and $$y$$ are called basic variables as they are determined once we make a choice of the free variable.

We will call this description of the solution space, in which the basic variables are written in terms of the free variables, a parametric description of the solution space.

2. Consider the matrix

\begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 0 \amp 4 \\ 0 \amp 1 \amp 0 \amp -3 \\ 0 \amp 0 \amp 1 \amp 1 \\ 0 \amp 0 \amp 0 \amp 0 \\ \end{array} \right]. \end{equation*}

The last equation gives

\begin{equation*} 0x +0y+0z = 0\text{,} \end{equation*}

which is true for any $$(x,y,z)\text{.}$$ We may safely ignore this equation since it does not provide a restriction on the choice of $$(x,y,z)\text{.}$$ We then see that there is a unique solution $$(x,y,z) = (4,-3,1)\text{.}$$

3. Consider the matrix

\begin{equation*} \left[ \begin{array}{rrr|r} 1 \amp 0 \amp 2 \amp 0 \\ 0 \amp 1 \amp -1 \amp 0 \\ 0 \amp 0 \amp 0 \amp 1 \\ \end{array} \right]. \end{equation*}

Beginning with the last equation, we see that

\begin{equation*} 0x +0y+0z = 0 = 1\text{,} \end{equation*}

which is not true for any $$(x,y,z)\text{.}$$ There is no solution to this particular equation and therefore no solution to the system of equations.

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