<p>Notice that for a consistent system the row-reduced augmented matrix has $n+1\in F$, so the largest element of $F$ does not refer to a variable. Also, for an inconsistent system, $n+1\in D$, and it then does not make much sense to discuss whether or not variables are free or dependent since there is no solution. Take a look back at <acroref type="definition" acro="IDV" /> and see why we did not need to consider the possibility of referencing $x_{n+1}$ as a dependent variable.</p>
-<p>With the characterization of <acroref type="theorem" acro="RCLS" />, we can explore the relationships between $r$ and $n$ in light of the consistency of a system of equations. First, a situation where we can quickly conclude the inconsistency of a system.</p>
-<theorem acro="ISRN" index="inconsistent linear systems">
-<title>Inconsistent Systems, $r$ and $n$</title>
-<p>Suppose $A$ is the augmented matrix of a system of linear equations in $n$ variables. Suppose also that $B$ is a row-equivalent matrix in reduced row-echelon form with $r$ rows that are not completely zeros. If $r=n+1$, then the system of equations is inconsistent.</p>
-<p>If $r=n+1$, then $D=\set{1,\,2,\,3,\,\ldots,\,n,\,n+1}$ and every column of $B$ contains a leading 1 and is a pivot column. In particular, the entry of column $n+1$ for row $r=n+1$ is a leading 1. <acroref type="theorem" acro="RCLS" /> then says that the system is inconsistent.</p>
-<p>Do not confuse <acroref type="theorem" acro="ISRN" /> with its converse! Go check out <acroref type="technique" acro="CV" /> right now.</p>
-<p>Next, if a system is consistent, we can distinguish between a unique solution and infinitely many solutions, and furthermore, we recognize that these are the only two possibilities.</p>
+<p>With the characterization of <acroref type="theorem" acro="RCLS" />, we can explore the relationships between $r$ and $n$ for a consistent system. We can distinguish between the case of a unique solution and infinitely many solutions, and furthermore, we recognize that these are the only two possibilities.</p>
<theorem acro="CSRN" index="consistent linear systems">
<title>Consistent Systems, $r$ and $n$</title>
-<p>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, <ie /> $r\leq n+1$. If $r=n+1$, then <acroref type="theorem" acro="ISRN" /> tells us that the system is inconsistent, contrary to our hypothesis. We are left with $r\leq n$.</p>
+<p>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, <ie /> $r\leq n+1$. If $r=n+1$, then every column of $B$ is a pivot column, and in particular, the last column is a pivot column. So <acroref type="theorem" acro="RCLS" /> tells us that the system is inconsistent, contrary to our hypothesis. We are left with $r\leq n$.</p>
<p>When $r=n$, we find $n-r=0$ free variables (<ie /> $F=\set{n+1}$) and any solution must equal the unique solution given by the first $n$ entries of column $n+1$ of $B$.</p>
<p>For each archetype that is a system of equations, the values of $n$ and $r$ are listed. Many also contain a few sample solutions. We can use this information profitably, as illustrated by four examples.
<ol><li> <acroref type="archetype" acro="A" /> has $n=3$ and $r=2$. It can be seen to be consistent by the sample solutions given. Its solution set then has $n-r=1$ free variables, and therefore will be infinite.
</li><li> <acroref type="archetype" acro="B" /> has $n=3$ and $r=3$. It can be seen to be consistent by the single sample solution given. Its solution set can then be described with $n-r=0$ free variables, and therefore will have just the single solution.
-</li><li> <acroref type="archetype" acro="H" /> has $n=2$ and $r=3$. In this case, $r=n+1$, so <acroref type="theorem" acro="ISRN" /> says the system is inconsistent. We should not try to apply <acroref type="theorem" acro="FVCS" /> to count free variables, since the theorem only applies to consistent systems. (What would happen if you did?)
-</li><li> <acroref type="archetype" acro="E" /> has $n=4$ and $r=3$. However, by looking at the reduced row-echelon form of the augmented matrix, we find a leading 1 in row 3, column 5. By <acroref type="theorem" acro="RCLS" /> we recognize the system as inconsistent. (Why doesn't this example contradict <acroref type="theorem" acro="ISRN" />?)
+</li><li> <acroref type="archetype" acro="H" /> has $n=2$ and $r=3$. In this case, column 3 must be a pivot column, so by <acroref type="theorem" acro="RCLS" />, the system is inconsistent. We should not try to apply <acroref type="theorem" acro="FVCS" /> to count free variables, since the theorem only applies to consistent systems. (What would happen if you did try to incorrectly apply <acroref type="theorem" acro="FVCS" />?)
+</li><li> <acroref type="archetype" acro="E" /> has $n=4$ and $r=3$. However, by looking at the reduced row-echelon form of the augmented matrix, we find that column 5 is a pivot column. By <acroref type="theorem" acro="RCLS" /> we recognize the system as inconsistent.
<![CDATA[0 & 0 & 0 & \leading{1}]]>
-Since $n = 3$ and $r = 4 = n+1$, <acroref type="theorem" acro="ISRN" /> guarantees that the system is inconsistent. Thus, we see that the given system has no solution.
+Since $n = 3$ and $r = 4 = n+1$, the last column is a pivot column and <acroref type="theorem" acro="RCLS" /> guarantees that the system is inconsistent. Thus, we see that the given system has no solution.
<exercise type="M" number="57" rough="Vars + equations + extra, what can be said??">
<problem contributor="robertbeezer">A system with 8 equations and 6 variables. The reduced row-echelon form of the augmented matrix of the system has 7 pivot columns.
-<solution contributor="robertbeezer">7 pivot columns implies that there are $r=7$ nonzero rows (so row 8 is all zeros in the reduced row-echelon form). Then $n+1=6+1=7=r$ and <acroref type="theorem" acro="ISRN" /> allows to conclude that the system is inconsistent.
+<solution contributor="robertbeezer">7 pivot columns implies that there are $r=7$ nonzero rows (so row 8 is all zeros in the reduced row-echelon form). The last column must be a pivot column and <acroref type="theorem" acro="RCLS" /> allows to conclude that the system is inconsistent.
+<exercise type="T" number="11" rough="r=n+1 is inconsistent">
+<problem contributor="robertbeezer">Suppose $A$ is the augmented matrix of a system of linear equations in $n$ variables. and that $B$ is a row-equivalent matrix in reduced row-echelon form with $r$ pivot columns. If $r=n+1$, prove that the system of equations is inconsistent.
+<solution contributor="robertbeezer">If $r=n+1$, then $D=\set{1,\,2,\,3,\,\ldots,\,n,\,n+1}$ and every column of $B$ is a pivot column. In particular, column $n+1$ is a pivot column. By <acroref type="theorem" acro="RCLS" /> the system is inconsistent.
<exercise type="T" number="20" rough="full generalization of TSS.M46, TSS.M70">
<problem contributor="manleyperkel">Suppose that $B$ is a matrix in reduced row-echelon form that is equivalent to the augmented matrix of a system of equations with $m$ equations in $n$ variables. Let $r$, $D$ and $F$ be as defined in <acroref type="definition" acro="RREF" />. Prove that $d_k\geq k$ for all $1\leq k\leq r$. Then suppose that $r\geq 2$ and <![CDATA[$1\leq k <\ell\leq r$]]> and determine what can you conclude, in general, about the following entries.