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Mathematical Expression Editor
Augmented Matrix Notation and Elementary Row Operations
Augmented Matrix Notation
Recall that the following three operations performed on a linear system are called
elementary row operations
Switching the order of two equations
Multiplying both sides of an equation by the same non-zero constant
Adding a multiple of one equation to another
In Introduction to Systems of Linear Equations we discussed why applying
elementary row operations to a linear system results in an equivalent system - a
linear system with the same solution set.
In this section we seek an efficient method for recording our computations as we
perform elementary row operations.
Consider the linear system \begin{equation} \label{eq:sys20originalsystem1} \begin{array}{ccccccccc} x &- &y&&&&&= &0 \\ 2x& -&2y&+&z&+&2w&=&4\\ & &y&&&+&w&=&0\\ & &&&2z&+&w&=&5 \end{array} \end{equation}
Our goal is to use elementary row operations to transform this system into an
equivalent system of the form \begin{equation} \begin{array}{ccccccccc} x & &&&&&&= &a \\ & &y&&&&&=&b\\ & &&&z&&&=&c\\ & &&&&&w&=&d \end{array} \end{equation}
We have to keep in mind that given an arbitrary system, an equivalent system of this
form may not exist (we will talk a lot more about this later). However, it does exist
in this case, and we would like to find a more efficient way of getting to it than
having to write and rewrite our equations at each step.
In this problem, we prompt you to perform elementary row operations on
(eq:sys20originalsystem1) and ask you to fill in the coefficients in the resulting equations. This is
a multi-step process. Steps will unfold automatically as you enter correct
answers.
ccess interactives through the online version of this text at
We start by subtracting twice row 1 from row 2. ()
Next, we add row 3 to row 1. ()
Subtract twice row 2 from row 4. ()
Divide row 4 by . ()
We will do three operations in one step.
We now exchange rows 2 and 3. () \begin{equation} \label{eq:sys20rref1} \begin{array}{ccccccccc} x &+ &0y&+&0z&+&0w&= &-1 \\ 0x& +&y&+&0z&+&0w&=&-1\\ 0x& +&0y&+&z&+&0w&=&2\\ 0x&+&0y&+&0z&+&w&=&1 \end{array} \end{equation}
If we drop all of the zero terms, we have: \begin{equation} \label{eq:sys20rrefnozeros} \begin{array}{ccccccccc} x & &&&&&&= &-1 \\ & &y&&&&&=&-1\\ & &&&z&&&=&2\\ &&&&&&w&=&1 \end{array} \end{equation}
Now we see that is the solution.
Observe that throughout the entire process, variables , , and remained in place; only
the coefficients in front of the variables and the entries on the right changed. Let’s try
to recreate this process without writing down the variables. We can capture the
original system in (eq:sys20originalsystem1) as follows:
The side to the left of the vertical bar is called the coefficient matrix, while the side
to the right of the bar is a vector that consists of constants on the right side of the
system. The coefficient matrix, together with the vector, is called an augmented
matrix.
We can capture all of the elementary row operations we performed earlier as
follows:
The last augmented matrix corresponds to systems in (eq:sys20rref1) and (eq:sys20rrefnozeros), and we can easily see
the solution.
Exploration init:augmentedmatrixex introduced us to some vocabulary terms. Let’s formalize our definitions.
Every linear system
can be written in the augmented matrix form as follows:
The array to the left of the vertical bar is called the coefficient matrix of the linear
system and is often given a capital letter name, like . The vertical array to the right
of the bar is called a constant vector.
We will sometimes use the following notation to represent an augmented
matrix.
The same elementary row operations that we perform on a system of equations can
be performed on the corresponding augmented matrix, or any matrix for that matter.
If a matrix can be obtained from another matrix by means of elementary row
operations, we say that the two matrices are row-equivalent.
Consider the
system
Recall that in Exploration init:augmentedmatrixex we converted the given system to an augmented matrix
form, then performed elementary row operations until we arrived at a “convenient”
form. We then converted the “convenient” augmented matrix back to a system of
equations and identified the solution. The term “convenient” is open to
interpretation. In this problem we will explore two “convenient” forms. Each one will
lead to a definition.
The augmented matrix in (eq:sys20reducedrowechelon) has the same convenient form as the one in (eq:sys20rref). This
augmented matrix corresponds to the system \begin{equation*} \begin{array}{ccccccc} x_1 & &&&&= &1/2 \\ & &x_2&&&=&-13/2\\ & &&&x_3&=&2 \end{array} \end{equation*}
This gives us the solution .
While the augmented matrix in (eq:sys20reducedrowechelon) was certainly “convenient”, we could have
converted back to the equation format a little earlier. Let’s take a look at the
augmented matrix in (eq:sys20rowechelon). Converting (eq:sys20rowechelon) to a system of equations gives us
\begin{equation*} \begin{array}{ccccccc} x_1 &- &x_2&-&3x_3&= &1\\ & &2x_2&+&5x_3&=&-3\\ & &&&x_3&=&2 \end{array} \end{equation*}
Substituting into the second equation and solving for gives us
Now substituting and into the first equation results in
This process is called back substitution and it produces the same solution as we
obtained earlier.
Observe that the coefficient matrices in (eq:sys20rref) and (eq:sys20reducedrowechelon) have the same format: 1’s along the
diagonal, zeros above and below the 1’s. The other “convenient” format, exhibited by
the coefficient matrix in (eq:sys20rowechelon), also has zeros below the diagonal, but not all of the
diagonal entries are 1’s and some of the entries above the diagonal are not zero. Each
of these formats gives rise to a definition. These definitions are the topic of the next
section.
Row-Echelon and Reduced Row-Echelon Forms
The first non-zero entry in a row of a matrix (when read from left to right) is called
the leading entry. When the leading entry is 1, we refer to it as a leading
1.
Row-Echelon Form A matrix is said to be in row-echelon form if:
(a)
All entries below each leading entry are 0.
(b)
Each leading entry is in a column to the right of the leading entries in the
rows above it.
(c)
All rows of zeros, if there are any, are located below non-zero rows.
The term row-echelon form can be applied to matrices whether or not they are
augmented matrices (matrices with the vertical bar). For example, both the
coefficient matrix and the augmented matrix in (eq:sys20rowechelon) are in row-echelon form.
Note that the leading entries form a staircase pattern. All entries below the
leading entries are zero, but the entries above the leading entries are not all
zero.
Below are two more examples of matrices in row-echelon form. The leading entries of
each matrix are boxed.
The difference between the coefficient matrix in (eq:sys20rowechelon) and the coefficient matrix in (eq:sys20reducedrowechelon) is
that the leading entries of the matrix in (eq:sys20reducedrowechelon) are all 1’s, and the matrix has zeros above
each leading 1. This motivates our next definition.
Reduced Row-Echelon Form A matrix that is already in row-echelon form is said to
be in reduced row-echelon form if:
(a)
Each leading entry is
(b)
All entries above and below each leading are
The following two matrices are in reduced row-echelon form. Note that there are ’ s
below and above each leading .
When solving linear systems using the augmented matrix notation, our goal will be to
transform the augmented matrix into a row-echelon or reduced row-echelon form.
The reduced row-echelon form of is denoted by . As we transform the augmented
matrix to its reduced row-echelon form, the coefficient matrix (the matrix to
the left of the bar) also gets transformed to its reduced row-echelon form,
.
Solve the system of equations or determine that the system is inconsistent.
We begin by rewriting the system in the augmented matrix form.
Our goal is to convert this matrix to its reduced row-echelon form by means of
elementary row operations. To do this, we will proceed from left to right and use
leading entries to wipe out all entries above and below them.
Our final matrix may not be quite as nice as the one in (eq:sys20reducedrowechelon), but it is in reduced
row-echelon form. Our next step is to convert our augmented matrix back to a
system of equations. We have:
We will rewrite the system as follows:
Now we see that we can assign any value to , then compute , and to obtain a
solution to the system. For example, let , then , and , so is a solution. If we let ,
then , and , so is also a solution. To capture all possibilities, we will let , where is
an arbitrary parameter.
We can think of the solution set in two different ways. First, the solution set is the
set of all points of the form
We can also think of the solution set in geometric terms by observing that \begin{align*} x&=-3+\frac{5}{3}t\\ y&=1-\frac{2}{3}t\\ z&=7-3t\\ w&=t \end{align*}
is a set of parametric equations that describes a line in . (See Formula form:paramlinend) This means
that the three hyperplanes given by the equations in the system intersect in a line,
producing infinitely many solutions to the system.
Observe that in Example ex:freevar1, variables , and correspond to the leading in the reduced
row-echelon. We say that , and are the leading variables. Variable is not a
leading variable; we refer to it as a free variable and assign a parameter to
it.
Solve the system of equations or determine that the system is inconsistent.
We rewrite the system in augmented matrix form and transform it to reduced
row-echelon form. We leave the details of the elementary row operations to the reader
and state the final result.
Converting back to a system of linear equations, we get
\begin{equation} \label{eq:sys20nosolutions}\begin{array}{ccccccc} 1x &+ &0y&-&(2/3)z&= &0 \\ 0x&+ &1y&-&(4/3)z&=&0\\ 0x&+ &0y&+&0z&=&1 \end{array} \end{equation}
The last equation in this system clearly has no solutions. We conclude that this
system (and the original system) is inconsistent.
Note that the last row of the reduced row-echelon form in (eq:sys20nosolutionsys) looks like
this
This row corresponds to the equation
which clearly has no solutions.
In general, if the reduced row-echelon form of the augmented matrix contains a
row
we can conclude that the system is inconsistent.
Solve the system of equations or determine that the system is inconsistent.
We rewrite the system in the augmented matrix form and transform it to reduced
row-echelon form. We leave the details of the elementary row operations to the reader
and state the final result.
Converting the augmented matrix back to a system of equations we get
Unlike the last equation in (eq:sys20nosolutions), the last equation in this system has infinitely many
solutions because all values of , and satisfy it. Since the last equation contributes
nothing, we will remove it and rewrite the system as
Variables and correspond to leading in the reduced row-echelon form. So, and are
the leading variables. Variable is a free variable. We let . Solutions to this system are
points of the form
We can also interpret the solutions as lying on a line in given by parametric
equations \begin{align*} x&=-\frac{9}{4}-\frac{5}{4}t\\ y&=-\frac{7}{4}-\frac{3}{4}t\\ z&=t\\ \end{align*}
This line is the line of intersection of the three planes.
Suppose a system of equations has the following reduced row-echelon form
What can you say about the system?
The system is inconsistentThe system
has infinitely many solutionsThe system has a unique solutionWe
would have to examine the original system to make the final determination