The Problem
In Strang’s Worked Example 3.2C, consider
\[ A= \begin{bmatrix} 1&2&1\\ 3&6&3\\ 4&8&c \end{bmatrix}. \]The goal is to see how the value of \(c\) can change the pivots, rank, free variables, and nullspace.
First Eliminate the Redundant Rows
Row 2 is three times row 1:
\[ R_2=3R_1. \]Therefore
\[ R_2\leftarrow R_2-3R_1 \]produces
\[ [0\;0\;0]. \]For row 3, subtract four times row 1:
\[ R_3\leftarrow R_3-4R_1. \]This gives
\[ [4\;8\;c]-4[1\;2\;1] = [0\;0\;c-4]. \]The whole example now depends on whether \(c-4\) is zero or nonzero.
Case 1: c Is Not 4
If
\[ c\neq4, \]then
\[ c-4\neq0. \]The entry \(c-4\) can therefore become a pivot.
To put the matrix in reduced row echelon form, divide its row by \(c-4\):
\[ [0\;0\;c-4] \longrightarrow [0\;0\;1]. \]This is valid because division by \(c-4\) is allowed when \(c\neq4\).
The reduced matrix becomes
\[ R= \begin{bmatrix} 1&2&0\\ 0&0&1\\ 0&0&0 \end{bmatrix}. \]What Is a Pivot?
In reduced row echelon form, a pivot is the leading \(1\) of a nonzero row.
Here the pivots are
\[ \begin{bmatrix} \boxed{1}&2&0\\ 0&0&\boxed{1}\\ 0&0&0 \end{bmatrix}. \]Therefore the pivot columns are columns 1 and 3.
What Is a Pivot Variable?
Each variable corresponds to one matrix column:
\[ \text{column 1}\leftrightarrow x_1, \qquad \text{column 2}\leftrightarrow x_2, \qquad \text{column 3}\leftrightarrow x_3. \]A pivot variable is a variable whose corresponding column contains a pivot.
Therefore:
\[ x_1\text{ and }x_3\text{ are pivot variables}. \]Column 2 contains no pivot, so
\[ \boxed{x_2\text{ is a free variable}.} \]Why the Number 2 Is Not a Pivot
In the first row
\[ [1\;2\;0], \]the leading nonzero entry is the \(1\) in column 1. That is the pivot.
The \(2\) in column 2 is not a new leading entry. In fact, column 2 is twice column 1 in the original matrix:
\[ a_2=2a_1. \]So column 2 adds no new independent direction and receives no pivot.
Pivot Entry Versus Pivot Variable
These are different ideas.
In reduced row echelon form:
\[ \boxed{\text{pivot entry}=1}. \]But a pivot variable is not necessarily equal to 1.
A pivot variable is simply the variable attached to a pivot column.
Its numerical value is normally determined by the free variables.
Solving When c Is Not 4
From
\[ Rx=0, \]we obtain
\[ x_1+2x_2=0, \] \[ x_3=0. \]Since \(x_2\) is free, let
\[ x_2=t. \]Then the pivot variables are determined:
\[ x_1=-2t, \qquad x_3=0. \]Setting the free variable to \(1\) gives the special solution
\[ \boxed{ s_1= \begin{bmatrix} -2\\1\\0 \end{bmatrix}} \]For \(c\neq4\), there are two pivots and one free variable:
\[ r=2, \qquad n-r=3-2=1. \]Case 2: c = 4
Now suppose
\[ c=4. \]Then
\[ c-4=0. \]So the row that previously could create a pivot becomes
\[ [0\;0\;0]. \]The reduced matrix is now
\[ R= \begin{bmatrix} 1&2&1\\ 0&0&0\\ 0&0&0 \end{bmatrix}. \]There is only one pivot, in column 1:
\[ \boxed{r=1}. \]Therefore \(x_1\) is the only pivot variable.
Both
\[ x_2\quad\text{and}\quad x_3 \]are free variables.
Two Free Variables Produce Two Special Solutions
The only equation is
\[ x_1+2x_2+x_3=0. \]Therefore
\[ x_1=-2x_2-x_3. \]First choose
\[ x_2=1, \qquad x_3=0. \]Then
\[ x_1=-2, \]giving
\[ s_1= \begin{bmatrix} -2\\1\\0 \end{bmatrix}. \]Next choose
\[ x_2=0, \qquad x_3=1. \]Then
\[ x_1=-1, \]giving
\[ s_2= \begin{bmatrix} -1\\0\\1 \end{bmatrix}. \]The Main Lesson
Changing only one value, \(c\), can change the structure of the matrix.
When \(c\neq4\):
\[ \boxed{r=2,\quad 1\text{ free variable},\quad 1\text{ special solution}.} \]When \(c=4\):
\[ \boxed{r=1,\quad 2\text{ free variables},\quad 2\text{ special solutions}.} \]The central relationship is
\[ \boxed{\text{fewer pivots}\Rightarrow\text{more free variables}\Rightarrow\text{larger nullspace}.} \]Mental Model
\[ \boxed{ \begin{array}{c} \text{pivot column }j\\ \Downarrow\\ x_j\text{ is a pivot variable}\\ \Downarrow\\ \text{its value is determined} \end{array}} \]while
\[ \boxed{ \begin{array}{c} \text{no pivot in column }j\\ \Downarrow\\ x_j\text{ is a free variable}\\ \Downarrow\\ \text{we are allowed to choose it} \end{array}} \]Do not confuse a pivot entry with a pivot variable. The pivot entry in RREF is \(1\), but the corresponding pivot variable can have many different values depending on the free variables.
Note metadata
- Note type: learning-note
- Subject: learning-notes
- Source: Introduction to Linear Algebra, 5th Edition — Worked Example 3.2C