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 equations 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 this 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.