A column containing a leading 1 will be called a
The set of column indices for the pivot columns will be denoted by $D=\set{d_1,\,d_2,\,d_3,\,\ldots,\,d_r}$ where while the columns that are not pivot columns will be denoted as $F=\set{f_1,\,f_2,\,f_3,\,\ldots,\,f_{nr}}$ where @@ 767,7 +767,7 @@ \end{bmatrix} This matrix has two zero rows and three leading 1's. So $r=3$. Columns 1, 5, and 6 are pivot columns, so $D=\set{1,\,5,\,6}$ and then $F=\set{2,\,3,\,4,\,7,\,8}$.
+This matrix has two zero rows and three leading 1's. So $r=3$. Columns 1, 5, and 6 are the three pivot columns, so $D=\set{1,\,5,\,6}$ and then $F=\set{2,\,3,\,4,\,7,\,8}$. @@ 1043,7 +1043,7 @@ We will now run through some examples of using these definitions and theorems to solve some systems of equations. From now on, when we have a matrix in reduced rowechelon form, we will mark the leading 1's with a small box. In your work, you can box 'em, circle 'em or write 'em in a different color
We will now run through some examples of using these definitions and theorems to solve some systems of equations. From now on, when we have a matrix in reduced rowechelon form, we will mark the leading 1's with a small box. This will help you count, and identify, the pivot columns. In your work, you can box 'em, circle 'em or write 'em in a different color
The number $r$ is the single most important piece of information we can get from the reduced rowechelon form of a matrix. It is defined as the number of nonzero rows, but since each nonzero row has a leading 1, it is also the number of leading 1's present. For each leading 1, we have a pivot column, so $r$ is also the number of pivot columns. Repeating ourselves, $r$ is the number of nonzero rows, the number of leading 1's and the number of pivot columns. Across different situations, each of these interpretations of the meaning of $r$ will be useful.
+The number $r$ is the single most important piece of information we can get from the reduced rowechelon form of a matrix. It is defined as the number of nonzero rows, but since each nonzero row has a leading 1, it is also the number of leading 1's present. For each leading 1, we have a pivot column, so $r$ is also the number of pivot columns. Repeating ourselves, $r$ is the number of nonzero rows, the number of leading 1's and the number of pivot columns. Across different situations, each of these interpretations of the meaning of $r$ will be useful, though it may be most helpful to think in terms of pivot columns.
Before proving some theorems about the possibilities for solution sets to systems of equations, let's analyze one particular system with an infinite solution set very carefully as an example. We'll use this technique frequently, and shortly we'll refine it slightly.
@@ 97,7 +97,7 @@ Let $i$ denote one of the $r=3$ nonzero rows, and then we see that we can solve the corresponding equation represented by this row for the variable $x_{d_i}$ and write it as 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). We'll do this now, but you can already see how the subscripts upon subscripts takes some getting used to. +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.Suppose $A$ is the augmented matrix of a consistent system of linear equations and $B$ is a rowequivalent matrix in reduced rowechelon form. Suppose $j$ is the index of a column of $B$ that contains the leading 1 for some row (
Suppose $A$ is the augmented matrix of a consistent system of linear equations and $B$ is a rowequivalent matrix in reduced rowechelon form. Suppose $j$ is the index of a pivot column of $B$. Then the variable $x_j$ is
There are leading 1's in columns 1, 3 and 4, so $D=\set{1,\,3,\,4}$. From this we know that the variables $x_1$, $x_3$ and $x_4$ will be dependent variables, and each of the $r=3$ nonzero rows of the rowreduced matrix will yield an expression for one of these three variables. The set $F$ is all the remaining column indices, $F=\set{2,\,5,\,6}$. The column index $6$ in $F$ means that the final column is not a pivot column, and thus the system is consistent (
Columns 1, 3 and 4 ae pivot columns, so $D=\set{1,\,3,\,4}$. From this we know that the variables $x_1$, $x_3$ and $x_4$ will be dependent variables, and each of the $r=3$ nonzero rows of the rowreduced matrix will yield an expression for one of these three variables. The set $F$ is all the remaining column indices, $F=\set{2,\,5,\,6}$. The column index $6$ in $F$ means that the final column is not a pivot column, and thus the system is consistent (
Suppose $A$ is the augmented matrix of a system of linear equations with $n$ variables. Suppose also that $B$ is a rowequivalent matrix in reduced rowechelon form with $r$ nonzero rows. Then the system of equations is inconsistent if and only if the leading 1 of row $r$ is located in column $n+1$ of $B$.
+Suppose $A$ is the augmented matrix of a system of linear equations with $n$ variables. Suppose also that $B$ is a rowequivalent matrix in reduced rowechelon form with $r$ nonzero rows. Then the system of equations is inconsistent if and only if column $n+1$ of $B$ is a pivot column.
If the leading 1 for row $r$ is located somewhere in columns 1 through $n$, then every preceding row's leading 1 is also located in columns 1 through $n$. In other words, since the last leading 1 is not in the last column, no leading 1 for any row is in the last column, due to the echelon layout of the leading 1's (
If column $n+1$ of $B$ is not a pivot column, the leading 1 for row $r$ is located somewhere in columns 1 through $n$. Then every preceding row's leading 1 is also located in columns 1 through $n$. In other words, since the last leading 1 is not in the last column, no leading 1 for any row is in the last column, due to the echelon layout of the leading 1's (aug.pivots()
. Since aug
has 5 columns, the final column is numbered 4
, which is present in the list of pivot columns, as we expect.
+We can look at the reduced rowechelon form of the augmented matrix and see a pivot column in the final column, so we know the system is inconsistent. However, we could just as easily not form the reduced rowechelon form and just look at the list of pivot columns computed by aug.pivots()
. Since aug
has 5 columns, the final column is numbered 4
, which is present in the list of pivot columns, as we expect.
One feature of Sage is that we can easily extend its capabilities by defining new commands. Here will create a function that checks if an augmented matrix represents a consistent system of equations. The syntax is just a bit complicated. lambda
is the word that indicates we are making a new function, the input is temporarily named A
(think A
ugmented), and the name of the function is consistent
. Everything following the colon will be evaluated and reported back as the output.
Suppose $A$ is the augmented matrix of a consistent system of linear equations with $n$ variables. Suppose also that $B$ is a rowequivalent matrix in reduced rowechelon form with $r$ rows that are not zero rows. Then $r\leq n$. If $r=n$, then the system has a unique solution,  then the system has infinitely many solutions.
+Suppose $A$ is the augmented matrix of a consistent system of linear equations with $n$ variables. Suppose also that $B$ is a rowequivalent matrix in reduced rowechelon form with $r$ pivot columns. Then $r\leq n$. If $r=n$, then the system has a unique solution, and if $rn$, then the system has infinitely many solutions.
This theorem contains three implications that we must establish. Notice first that $B$ has $n+1$ columns, so there can be at most $n+1$ pivot columns,
When $r=n$, we find $nr=0$ free variables (
When $r=n$, we find $nr=0$ free variables (

we have $nr>0$ free variables,
corresponding to columns of $B$ without a leading 1, excepting the final column, which also does not contain a leading 1 by
When $rn$, we have $nr>0$ free variables. Choose one free variable and set all the other free variables to zero. Now, set the chosen free variable to any fixed value. It is possible to then determine the values of the dependent variables to create a solution to the system. By setting the chosen free variable to different values, in this manner we can create infinitely many solutions.