Consider a linear system of $m=5$ equations in $n=7$ variables, having the augmented matrix $A$. \begin{equation*} A= \begin{bmatrix} 2 & 1 & -1 & -2 & 2 & 1 & 5 & 21 \\ 1 & 1 & -3 & 1 & 1 & 1 & 2 & -5 \\ 1 & 2 & -8 & 5 & 1 & 1 & -6 & -15 \\ 3 & 3 & -9 & 3 & 6 & 5 & 2 & -24 \\ -2 & -1 & 1 & 2 & 1 & 1 & -9 & -30 \end{bmatrix} \end{equation*}
Row-reducing we obtain the matrix \begin{equation*} B= \begin{bmatrix} \leading{1} & 0 & 2 & -3 & 0 & 0 & 9 & 15 \\ 0 & \leading{1} & -5 & 4 & 0 & 0 & -8 & -10 \\ 0 & 0 & 0 & 0 & \leading{1} & 0 & -6 & 11 \\ 0 & 0 & 0 & 0 & 0 & \leading{1} & 7 & -21 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} \end{equation*} and we see $r=4$ pivot columns. Also, $D=\set{1,\,2,\,5,\,6}$ so the dependent variables are then $x_1,\,x_2,\,x_5,$ and $x_6$. $F=\set{3,\,4,\,7,\,8}$ so the $n-r=3$ free variables are $x_3,\,x_4$ and $x_7$. We will express a generic solution for the system by two different methods: both a decomposition and a construction.
First, we will decompose (Proof Technique DC) a solution vector. Rearranging each equation represented in the row-reduced form of the augmented matrix by solving for the dependent variable in each row yields the vector equality, \begin{align*} \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}&= \colvector{ 15-2x_3+3x_4-9x_7\\ -10+5x_3-4x_4+8x_7\\ x_3\\ x_4\\ 11+6x_7\\ -21-7x_7\\ x_7 } \end{align*} Now we will use the definitions of column vector addition and scalar multiplication to decompose this generic solution vector as a linear combination, \begin{align*} &= \colvector{15\\ -10\\ 0\\ 0\\ 11\\ -21\\ 0 } + \colvector{ -2x_3\\ 5x_3\\ x_3\\ 0\\ 0\\ 0\\ 0 } + \colvector{ 3x_4\\ -4x_4\\ 0\\ x_4\\ 0\\ 0\\ 0 } + \colvector{ -9x_7\\ 8x_7\\ 0\\ 0\\ 6x_7\\ -7x_7\\ x_7 } &&\knowl{./knowls/definition.CVA.knowl}{\text{Definition CVA}}\\ &= \colvector{15\\ -10\\ 0\\ 0\\ 11\\ -21\\ 0 } + x_3\colvector{ -2\\ 5\\ 1\\ 0\\ 0\\ 0\\ 0 } + x_4\colvector{ 3\\ -4\\ 0\\ 1\\ 0\\ 0\\ 0 } + x_7\colvector{ -9\\ 8\\ 0\\ 0\\ 6\\ -7\\ 1 } &&\knowl{./knowls/definition.CVSM.knowl}{\text{Definition CVSM}} \end{align*} We will now develop the same linear combination a bit quicker, using three steps. While the method above is instructive, the method below will be our preferred approach.
Step 1. Write the vector of variables as a fixed vector, plus a linear combination of $n-r$ vectors, using the free variables as the scalars. \begin{equation*} \vect{x}= \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}= \colvector{\ \\\ \\\ \\\ \\\ \\\ \\\ }+x_3\colvector{\ \\\ \\\ \\\ \\\ \\\ \\\ }+x_4\colvector{\ \\\ \\\ \\\ \\\ \\\ \\\ }+x_7\colvector{\ \\\ \\\ \\\ \\\ \\\ \\\ } \end{equation*}
Step 2. Use 0's and 1's to ensure equality for the entries of the vectors with indices in $F$ (corresponding to the free variables). \begin{equation*} \vect{x}= \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}= \colvector{\\ \\ 0\\ 0\\ \\ \\ 0}+x_3\colvector{\\ \\ 1\\ 0\\ \\ \\ 0}+x_4\colvector{\\ \\ 0\\ 1\\ \\ \\ 0}+x_7\colvector{\\ \\ 0\\ 0\\ \\ \\ 1} \end{equation*}
Step 3. For each dependent variable, use the augmented matrix to formulate an equation expressing the dependent variable as a constant plus multiples of the free variables. Convert this equation into entries of the vectors that ensure equality for each dependent variable, one at a time. \begin{align*} x_1&=15-2x_3+3x_4-9x_7\ \Rightarrow\\&\vect{x}= \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}= \colvector{15\\ \\ 0\\ 0\\ \\ \\ 0}+ x_3\colvector{-2\\ \\ 1\\ 0\\ \\ \\ 0}+ x_4\colvector{3\\ \\ 0\\ 1\\ \\ \\ 0}+ x_7\colvector{-9\\ \\ 0\\ 0\\ \\ \\ 1}\\ x_2&=-10+5x_3-4x_4+8x_7\ \Rightarrow\\&\vect{x}= \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}= \colvector{15\\ -10\\ 0\\ 0\\ \\ \\ 0}+ x_3\colvector{-2\\ 5\\ 1\\ 0\\ \\ \\ 0}+ x_4\colvector{3\\ -4\\ 0\\ 1\\ \\ \\ 0}+ x_7\colvector{-9\\ 8\\ 0\\ 0\\ \\ \\ 1}\\ x_5&=11+6x_7\ \Rightarrow\\&\vect{x}= \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}= \colvector{15\\ -10\\ 0\\ 0\\ 11\\ \\ 0}+ x_3\colvector{-2\\ 5\\ 1\\ 0\\ 0\\ \\ 0}+ x_4\colvector{3\\ -4\\ 0\\ 1\\ 0\\ \\ 0}+ x_7\colvector{-9\\ 8\\ 0\\ 0\\ 6\\ \\ 1}\\ x_6&=-21-7x_7\ \Rightarrow\\&\vect{x}= \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}= \colvector{15\\ -10\\ 0\\ 0\\ 11\\ -21\\ 0}+ x_3\colvector{-2\\ 5\\ 1\\ 0\\ 0\\ 0\\ 0}+ x_4\colvector{3\\ -4\\ 0\\ 1\\ 0\\ 0\\ 0}+ x_7\colvector{-9\\ 8\\ 0\\ 0\\ 6\\ -7\\ 1} \end{align*}
This final form of a typical solution is especially pleasing and useful. For example, we can build solutions quickly by choosing values for our free variables, and then compute a linear combination. For example \begin{align*} x_3&=2,\, x_4=-4,\, x_7=3 \quad\quad\Rightarrow\\ \vect{x}&= \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}= \colvector{15\\ -10\\ 0\\ 0\\ 11\\ -21\\ 0}+ (2)\colvector{-2\\ 5\\ 1\\ 0\\ 0\\ 0\\ 0}+ (-4)\colvector{3\\ -4\\ 0\\ 1\\ 0\\ 0\\ 0}+ (3)\colvector{-9\\ 8\\ 0\\ 0\\ 6\\ -7\\ 1} = \colvector{-28\\40\\2\\-4\\29\\-42\\3} \end{align*} or perhaps, \begin{align*} x_3&=5,\, x_4=2,\, x_7=1 \quad\quad\Rightarrow\\ \vect{x}&= \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}= \colvector{15\\ -10\\ 0\\ 0\\ 11\\ -21\\ 0}+ (5)\colvector{-2\\ 5\\ 1\\ 0\\ 0\\ 0\\ 0}+ (2)\colvector{3\\ -4\\ 0\\ 1\\ 0\\ 0\\ 0}+ (1)\colvector{-9\\ 8\\ 0\\ 0\\ 6\\ -7\\ 1} = \colvector{2\\15\\5\\2\\17\\-28\\1} \end{align*} or even, \begin{align*} x_3&=0,\, x_4=0,\, x_7=0 \quad\quad\Rightarrow\\ \vect{x}&= \colvector{x_1\\x_2\\x_3\\x_4\\x_5\\x_6\\x_7}= \colvector{15\\ -10\\ 0\\ 0\\ 11\\ -21\\ 0}+ (0)\colvector{-2\\ 5\\ 1\\ 0\\ 0\\ 0\\ 0}+ (0)\colvector{3\\ -4\\ 0\\ 1\\ 0\\ 0\\ 0}+ (0)\colvector{-9\\ 8\\ 0\\ 0\\ 6\\ -7\\ 1} = \colvector{15\\ -10\\ 0\\ 0\\ 11\\ -21\\ 0} \end{align*}
So we can compactly express all of the solutions to this linear system with just 4 fixed vectors, provided we agree how to combine them in a linear combinations to create solution vectors.
Suppose you were told that the vector $\vect{w}$ below was a solution to this system of equations. Could you turn the problem around and write $\vect{w}$ as a linear combination of the four vectors $\vect{c}$, $\vect{u}_1$, $\vect{u}_2$, $\vect{u}_3$? (See Exercise LC.M11.) \begin{align*} \vect{w}&=\colvector{100\\-75\\7\\9\\-37\\35\\-8} & \vect{c}&=\colvector{15\\ -10\\ 0\\ 0\\ 11\\ -21\\ 0} & \vect{u}_1&=\colvector{-2\\ 5\\ 1\\ 0\\ 0\\ 0\\ 0} & \vect{u}_2&=\colvector{3\\ -4\\ 0\\ 1\\ 0\\ 0\\ 0} & \vect{u}_3&=\colvector{-9\\ 8\\ 0\\ 0\\ 6\\ -7\\ 1} \end{align*}