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Guias de estudo > College Algebra

Row Operations and The Augmented Matrix

Learning Objectives

  • Write the augmented matrix for a system of equations
  • Perform row operations on an augmented matrix
A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix. For example, consider the following 2×22\times 2 system of equations.
3x+4y=74x−2y=5\begin{array}{l}3x+4y=7\\ 4x - 2y=5\end{array}
We can write this system as an augmented matrix:
[344−2 ∣ 75]\left[\begin{array}{rr}\hfill 3& \hfill 4\\ \hfill 4& \hfill -2\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 7\\ \hfill 5\end{array}\right]
We can also write a matrix containing just the coefficients. This is called the coefficient matrix.
[344−2]\left[\begin{array}{cc}3& 4\\ 4& -2\end{array}\right]
A three-by-three system of equations such as
3x−y−z=0 x+y=5 2x−3z=2\begin{array}{l}3x-y-z=0\hfill \\ \text{ }x+y=5\hfill \\ \text{ }2x - 3z=2\hfill \end{array}
has a coefficient matrix
[3−1−111020−3]\left[\begin{array}{rrr}\hfill 3& \hfill -1& \hfill -1\\ \hfill 1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 0& \hfill -3\end{array}\right]
and is represented by the augmented matrix
[3−1−111020−3 ∣ 052]\left[\begin{array}{rrr}\hfill 3& \hfill -1& \hfill -1\\ \hfill 1& \hfill 1& \hfill 0\\ \hfill 2& \hfill 0& \hfill -3\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 0\\ \hfill 5\\ \hfill 2\end{array}\right]
Notice that the matrix is written so that the variables line up in their own columns: x-terms go in the first column, y-terms in the second column, and z-terms in the third column. It is very important that each equation is written in standard form ax+by+cz=dax+by+cz=d so that the variables line up. When there is a missing variable term in an equation, the coefficient is 0.

How To: Given a system of equations, write an augmented matrix.

  1. Write the coefficients of the x-terms as the numbers down the first column.
  2. Write the coefficients of the y-terms as the numbers down the second column.
  3. If there are z-terms, write the coefficients as the numbers down the third column.
  4. Draw a vertical line and write the constants to the right of the line.

Example: Writing the Augmented Matrix for a System of Equations

Write the augmented matrix for the given system of equations.
 x+2y−z=3 2x−y+2z=6 x−3y+3z=4\begin{array}{l}\text{ }x+2y-z=3\hfill \\ \text{ }2x-y+2z=6\hfill \\ \text{ }x - 3y+3z=4\hfill \end{array}

Answer: The augmented matrix displays the coefficients of the variables, and an additional column for the constants.

[12−12−121−33 ∣ 364]\left[\begin{array}{rrr}\hfill 1& \hfill 2& \hfill -1\\ \hfill 2& \hfill -1& \hfill 2\\ \hfill 1& \hfill -3& \hfill 3\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 3\\ \hfill 6\\ \hfill 4\end{array}\right]

Try It

Write the augmented matrix of the given system of equations.
4x−3y=113x+2y=4\begin{array}{l}4x - 3y=11\\ 3x+2y=4\end{array}

Answer: A+B=[2116   0−3]+[31−4−253]=[2+31+11+(−4)6+(−2)0+5−3+3]=[52−3450]A+B=\left[\begin{array}{c}2\\ 1\\ 1\end{array}\begin{array}{c}6\\ \text{ }\text{ }\text{ }0\\ -3\end{array}\right]+\left[\begin{array}{c}3\\ 1\\ -4\end{array}\begin{array}{c}-2\\ 5\\ 3\end{array}\right]=\left[\begin{array}{c}2+3\\ 1+1\\ 1+\left(-4\right)\end{array}\begin{array}{c}6+\left(-2\right)\\ 0+5\\ -3+3\end{array}\right]=\left[\begin{array}{c}5\\ 2\\ -3\end{array}\begin{array}{c}4\\ 5\\ 0\end{array}\right]

Row Operations

Now that we can write systems of equations in augmented matrix form, we will examine the various row operations that can be performed on a matrix, such as addition, multiplication by a constant, and interchanging rows. Performing row operations on a matrix is the method we use for solving a system of equations. In order to solve the system of equations, we want to convert the matrix to row-echelon form, in which there are ones down the main diagonal from the upper left corner to the lower right corner, and zeros in every position below the main diagonal as shown.
Row-echelon form[1ab01d001]\begin{array}{c}\text{Row-echelon form}\\ \left[\begin{array}{ccc}1& a& b\\ 0& 1& d\\ 0& 0& 1\end{array}\right]\end{array}
We use row operations corresponding to equation operations to obtain a new matrix that is row-equivalent in a simpler form. Here are the guidelines to obtaining row-echelon form.
  1. In any nonzero row, the first nonzero number is a 1. It is called a leading 1.
  2. Any all-zero rows are placed at the bottom on the matrix.
  3. Any leading 1 is below and to the right of a previous leading 1.
  4. Any column containing a leading 1 has zeros in all other positions in the column.
To solve a system of equations we can perform the following row operations to convert the coefficient matrix to row-echelon form and do back-substitution to find the solution.
  1. Interchange rows. (Notation: Ri↔Rj{R}_{i}\leftrightarrow {R}_{j} )
  2. Multiply a row by a constant. (Notation: cRic{R}_{i} )
  3. Add the product of a row multiplied by a constant to another row. (Notation: Ri+cRj{R}_{i}+c{R}_{j})
Each of the row operations corresponds to the operations we have already learned to solve systems of equations in three variables. With these operations, there are some key moves that will quickly achieve the goal of writing a matrix in row-echelon form. To obtain a matrix in row-echelon form for finding solutions, we use Gaussian elimination, a method that uses row operations to obtain a 1 as the first entry so that row 1 can be used to convert the remaining rows.

A General Note: Gaussian Elimination

The Gaussian elimination method refers to a strategy used to obtain the row-echelon form of a matrix. The goal is to write matrix AA with the number 1 as the entry down the main diagonal and have all zeros below.
A=[a11a12a13a21a22a23a31a32a33]→After Gaussian eliminationA=[1b12b1301b23001]A=\left[\begin{array}{rrr}\hfill {a}_{11}& \hfill {a}_{12}& \hfill {a}_{13}\\ \hfill {a}_{21}& \hfill {a}_{22}& \hfill {a}_{23}\\ \hfill {a}_{31}& \hfill {a}_{32}& \hfill {a}_{33}\end{array}\right]\stackrel{\text{After Gaussian elimination}}{\to }A=\left[\begin{array}{rrr}\hfill 1& \hfill {b}_{12}& \hfill {b}_{13}\\ \hfill 0& \hfill 1& \hfill {b}_{23}\\ \hfill 0& \hfill 0& \hfill 1\end{array}\right]
The first step of the Gaussian strategy includes obtaining a 1 as the first entry, so that row 1 may be used to alter the rows below.

How To: Given an augmented matrix, perform row operations to achieve row-echelon form.

  1. The first equation should have a leading coefficient of 1. Interchange rows or multiply by a constant, if necessary.
  2. Use row operations to obtain zeros down the first column below the first entry of 1.
  3. Use row operations to obtain a 1 in row 2, column 2.
  4. Use row operations to obtain zeros down column 2, below the entry of 1.
  5. Use row operations to obtain a 1 in row 3, column 3.
  6. Continue this process for all rows until there is a 1 in every entry down the main diagonal and there are only zeros below.
  7. If any rows contain all zeros, place them at the bottom.

Example: Performing Row Operations on a 3×3 Augmented Matrix to Obtain Row-Echelon Form

Perform row operations on the given matrix to obtain row-echelon form.
[1−342−56−334 ∣ 366]\left[\begin{array}{rrr}\hfill 1& \hfill -3& \hfill 4\\ \hfill 2& \hfill -5& \hfill 6\\ \hfill -3& \hfill 3& \hfill 4\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 3\\ \hfill 6\\ \hfill 6\end{array}\right]

Answer: The first row already has a 1 in row 1, column 1. The next step is to multiply row 1 by −2-2 and add it to row 2. Then replace row 2 with the result.

−2R1+R2=R2→[1−3401−2−334∣306]-2{R}_{1}+{R}_{2}={R}_{2}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill -3& \hfill & \hfill 3& \hfill & \hfill 4& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 6\end{array}\right]
Next, obtain a zero in row 3, column 1.
3R1+R3=R3→[1−3401−20−616∣3015]3{R}_{1}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill -6& \hfill & \hfill 16& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 15\end{array}\right]
Next, obtain a zero in row 3, column 2.
6R2+R3=R3→[1−3401−2004∣3015]6{R}_{2}+{R}_{3}={R}_{3}\to \left[\begin{array}{rrrrrr}\hfill 1& \hfill & \hfill -3& \hfill & \hfill 4& \hfill \\ \hfill 0& \hfill & \hfill 1& \hfill & \hfill -2& \hfill \\ \hfill 0& \hfill & \hfill 0& \hfill & \hfill 4& \hfill \end{array}|\begin{array}{rr}\hfill & \hfill 3\\ \hfill & \hfill 0\\ \hfill & \hfill 15\end{array}\right]
The last step is to obtain a 1 in row 3, column 3.
12R3=R3→[1−3401−2001 ∣ 3−6212]\frac{1}{2}{R}_{3}={R}_{3}\to \left[\begin{array}{rrr}\hfill 1& \hfill -3& \hfill 4\\ \hfill 0& \hfill 1& \hfill -2\\ \hfill 0& \hfill 0& \hfill 1\end{array}\text{ }|\text{ }\begin{array}{r}\hfill 3\\ \hfill -6\\ \hfill \frac{21}{2}\end{array}\right]

Try It

Write the system of equations in row-echelon form.
 x−2y+3z=9 −x+3y=−42x−5y+5z=17\begin{array}{l}\text{ }x - 2y+3z=9\hfill \\ \text{ }-x+3y=-4\hfill \\ 2x - 5y+5z=17\hfill \end{array}

Answer: [1−5252 015001∣17292]\left[\begin{array}{ccc}1& -\frac{5}{2}& \frac{5}{2}\\ \text{ }0& 1& 5\\ 0& 0& 1\end{array}|\begin{array}{c}\frac{17}{2}\\ 9\\ 2\end{array}\right]

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