Archetype I is the system of $m=4$ equations in $n=7$ variables. \begin{align*} x_1 +4x_2 - x_4 + 7x_6 - 9x_7 &= 3\\ 2x_1 + 8x_2 - x_3 + 3x_4 + 9x_5 - 13x_6 + 7x_7 &= 9\\ 2x_3 -3x_4 -4x_5 +12x_6 -8x_7 &= 1\\ -x_1 - 4x_2 + 2x_3 +4x_4 + 8x_5 - 31x_6 + 37x_7 &= 4 \end{align*} This system has a $4\times 8$ augmented matrix that is row-equivalent to the following matrix (check this!), and which is in reduced row-echelon form (the existence of this matrix is guaranteed by Theorem REMEF and its uniqueness is guaranteed by Theorem RREFU), \begin{align*} \begin{bmatrix} \leading{1} & 4 & 0 & 0 & 2 & 1 & -3 & 4 \\ 0 & 0 & \leading{1} & 0 & 1 & -3 & 5 & 2 \\ 0 & 0 & 0 & \leading{1} & 2 & -6 & 6 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{align*}
So we find that $r=3$ and \begin{align*} D&=\set{d_1,\,d_2,\,d_3}=\set{1,\,3,\,4} & F&=\set{f_1,\,f_2,\,f_3,\,f_4,\,f_5}=\set{2,\,5,\,6,\,7,\,8} \end{align*} Let $i$ denote any one of the $r=3$ nonzero rows. Then the index $d_i$ is a pivot column. It will be easy in this case to use the equation represented by row $i$ to write an expression for the variable $x_{d_i}$. It will be a linear function of the variables $x_{f_1},\,x_{f_2},\,x_{f_3},\,x_{f_4}$ (notice that $f_5=8$ does not reference a variable, but does tell us the final column is not a pivot column). We will now construct these three expressions. Notice that using subscripts upon subscripts takes some getting used to. \begin{align*} (i=1)& & x_{d_1}&=x_1=4-4x_2-2x_5-x_6+3x_7\\ (i=2)& & x_{d_2}&=x_3=2-x_5+3x_6-5x_7\\ (i=3)& & x_{d_3}&=x_4=1-2x_5+6x_6-6x_7 \end{align*}
Each element of the set $F=\set{f_1,\,f_2,\,f_3,\,f_4,\,f_5}=\set{2,\,5,\,6,\,7,\,8}$ is the index of a variable, except for $f_5=8$. We refer to $x_{f_1}=x_2$, $x_{f_2}=x_5$, $x_{f_3}=x_6$ and $x_{f_4}=x_7$ as “free” (or “independent”) variables since they are allowed to assume any possible combination of values that we can imagine and we can continue on to build a solution to the system by solving individual equations for the values of the other (“dependent”) variables.
Each element of the set $D=\set{d_1,\,d_2,\,d_3}=\set{1,\,3,\,4}$ is the index of a variable. We refer to the variables $x_{d_1}=x_1$, $x_{d_2}=x_3$ and $x_{d_3}=x_4$ as “dependent” variables since they depend on the independent variables. More precisely, for each possible choice of values for the independent variables we get exactly one set of values for the dependent variables that combine to form a solution of the system.
To express the solutions as a set, we write \begin{equation*} \setparts{ \colvector{ 4-4x_2-2x_5-x_6+3x_7\\ x_2\\ 2-x_5+3x_6-5x_7\\ 1-2x_5+6x_6-6x_7\\ x_5\\ x_6\\ x_7 } }{ x_2,\,x_5,\,x_6,\,x_7\in\complex{\null} } \end{equation*}
The condition that $x_2,\,x_5,\,x_6,\,x_7\in\complex{\null}$ is how we specify that the variables $x_2,\,x_5,\,x_6,\,x_7$ are “free” to assume any possible values.
This systematic approach to solving a system of equations will allow us to create a precise description of the solution set for any consistent system once we have found the reduced row-echelon form of the augmented matrix. It will work just as well when the set of free variables is empty and we get just a single solution. And we could program a computer to do it! Now have a whack at Archetype J (Exercise TSS.C10), mimicking the discussion in this example. We'll still be here when you get back.