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

https://ximera.osu.edu/linearalgebradzv3/LinearAlgebraInteractiveIntro.

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:

\begin{equation} \label{eq:sys20rref}\left [\begin{array}{cccc|c} 1&0&0&0&-1\\0&1&0&0&-1\\0&0&1&0&2\\0&0&0&1&1 \end{array}\right ] \end{equation}

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.

\begin{align} &\left [\begin{array}{ccc|c} 1&-1&-3&1\\4&-2&-7&1\\-1&1&4&1 \end{array}\right ]\nonumber \\ \begin{array}{c} \\ \xrightarrow{R_2-4R_1}\\ \xrightarrow{R_3+R_1}\\ \end{array} &\left [\begin{array}{ccc|c} 1&-1&-3&1\\0&2&5&-3\\0&0&1&2 \end{array}\right ]\label{eq:sys20rowechelon}\\ \begin{array}{c} \\ \xrightarrow{(1/2)R_2}\\ \\ \end{array} &\left [\begin{array}{ccc|c} 1&-1&-3&1\\0&1&5/2&-3/2\\0&0&1&2 \end{array}\right ]\nonumber \\ \begin{array}{c} \xrightarrow{R_1+R_2}\\ \\ \\ \end{array}&\left [\begin{array}{ccc|c} 1&0&-1/2&-1/2\\0&1&5/2&-3/2\\0&0&1&2 \end{array}\right ]\nonumber \\ \begin{array}{c} \xrightarrow{R_1+ (1/2)R_3}\\ \xrightarrow{R_2- (5/2)R_3}\\ \\ \end{array} &\left [\begin{array}{ccc|c} 1&0&0&1/2\\0&1&0&-13/2\\0&0&1&2 \end{array}\right ]\label{eq:sys20reducedrowechelon} \end{align}

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 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.

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, .

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.

Practice Problems

Problems prob:rrefmultchoice1-prob:rrefmultchoice5

Determine whether each augmented matrix shown below is in reduced row-echelon form.

Yes No
Yes No
Yes No
Yes No
Yes No
Fill in the steps that lead to the reduced row-echelon form in Example ex:nosolutionssys.
Fill in the steps that lead to the reduced row-echelon form in Example ex:rrefinfmanysolutionssys.
Suppose a system of equations has the following reduced row-echelon form What can you say about the system?
The system is inconsistent The system has infinitely many solutions The system has a unique solution We would have to examine the original system to make the final determination

Problems prob:sys20solvesys1-prob:sys20solvesys2

Solve each system of equations.

Solution:

Solution: (Enter your answers in fraction form)

Example Source

Linear system in Exploration init:augmentedmatrixex comes from Jim Hefferon’s Linear Algebra. (CC-BY-NC-SA)

Jim Hefferon, Linear Algebra, 3rd edition, page 6.

2024-09-26 20:17:34