Row Operations and Augmented Matrices

Augmented Matrices and Row Operations

Learning Outcomes

  • A matrix can represent and solve a system of equations.

  • Express a system in matrix form by extracting coefficients and constants.

  • Use a vertical line to separate coefficients from constants, creating an augmented matrix.

  • Write the augmented matrix for a system of equations.

  • Perform row operations on an augmented matrix.

Augmented Matrix

  • A matrix form of a system of equations.

  • Coefficients of variables and constants become entries of the matrix.

  • A vertical line separates coefficient entries from constants (replaces equal signs).

Example: 2x2 System
  • System of equations:
    3x+4y=73x + 4y = 7
    4x2y=54x - 2y = 5

  • Augmented matrix:
    [3amp;4amp;7 4amp;2amp;5]\begin{bmatrix} 3 & 4 & 7 \ 4 & -2 & 5 \end{bmatrix}

Coefficient Matrix
  • A matrix containing only the coefficients of the variables.

  • Example:
    [3amp;4 4amp;2]\begin{bmatrix} 3 & 4 \ 4 & -2 \end{bmatrix}

Example: 3x3 System
  • System of equations:
    3xyz=03x - y - z = 0
    x+y=5x + y = 5
    2x3z=22x - 3z = 2

  • Coefficient matrix:

    [3amp;1amp;11amp;1amp;02amp;0amp;3]\begin{bmatrix} 3 & -1 & -1 \\ 1 & 1 & 0 \\ 2 & 0 & -3 \end{bmatrix}

  • Augmented matrix:
    [3amp;1amp;1amp;0 1amp;1amp;0amp;5 2amp;0amp;3amp;2]\begin{bmatrix} 3 & -1 & -1 & 0 \ 1 & 1 & 0 & 5 \ 2 & 0 & -3 & 2 \end{bmatrix} y, z terms).

  • Equations must be in standard form: ax+by+cz=dax + by + cz = d

  • Missing variable terms have a coefficient of 0.

How To: Write an Augmented Matrix

  1. Write coefficients of x-terms down the first column.

  2. Write coefficients of y-terms down the second column.

  3. If there are z-terms, write coefficients down the third column.

  4. Draw a vertical line and write the constants to the right.

Example: Writing the Augmented Matrix
  • System of equations:
    x+2yz=3x + 2y - z = 3
    2xy+2z=62x - y + 2z = 6
    x3y+3z=4x - 3y + 3z = 4

  • Augmented matrix:
    [1amp;2amp;1amp;3 2amp;1amp;2amp;6 1amp;3amp;3amp;4]\begin{bmatrix} 1 & 2 & -1 & 3 \ 2 & -1 & 2 & 6 \ 1 & -3 & 3 & 4 \end{bmatrix}

Try It
  • System of equations:
    4x3y=114x - 3y = 11
    3x+2y=43x + 2y = 4

Writing a System of Equations from an Augmented Matrix

  • Important to move between formats for smoother solutions.

  • Use the information in an augmented matrix to write the system of equations in standard form.

Example: Writing a System of Equations
  • Augmented matrix:
    [1amp;3amp;5amp;2 2amp;5amp;4amp;5 3amp;5amp;4amp;6]\begin{bmatrix} 1 & -3 & -5 & -2 \ 2 & -5 & -4 & 5 \ -3 & 5 & 4 & 6 \end{bmatrix}

  • System of equations:
    x3y5z=2x - 3y - 5z = -2
    2x5y4z=52x - 5y - 4z = 5
    3x+5y+4z=6-3x + 5y + 4z = 6

Try It
  • Augmented matrix:
    [1amp;1amp;1amp;5 2amp;1amp;3amp;1 0amp;1amp;1amp;9]\begin{bmatrix} 1 & -1 & 1 & 5 \ 2 & -1 & 3 & 1 \ 0 & 1 & 1 & -9 \end{bmatrix}

Row Operations

  • Row operations: addition, multiplication by a constant, interchanging rows.

  • Used to solve systems of equations.

  • Convert the matrix to row-echelon form.

Row-Echelon Form
  • Ones down the main diagonal (upper left to lower right).

  • Zeros in every position below the main diagonal.

  • Example:
    [1amp;aamp;b 0amp;1amp;d 0amp;0amp;1]\begin{bmatrix} 1 & a & b \ 0 & 1 & d \ 0 & 0 & 1 \end{bmatrix}

Guidelines for Obtaining Row-Echelon Form
  1. First nonzero number in any nonzero row is a 1 (leading 1).

  2. All-zero rows are at the bottom.

  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.

Row Operations Notation
  1. Interchange rows: RiRjRi \leftrightarrow Rj

  2. Multiply a row by a constant: cRicR_i

  3. Add the product of a row multiplied by a constant to another row: Ri+cRjRi + cRj

Gaussian Elimination

  • Method to obtain row-echelon form.

  • Goal: Write matrix A with 1s down the main diagonal and zeros below.

  • A=[a11amp;a12amp;a13 a21amp;a22amp;a23 a31amp;a32amp;a33]After Gaussian eliminationA=[1amp;b12amp;b13 0amp;1amp;b23 0amp;0amp;1]A = \begin{bmatrix} a{11} & a{12} & a{13} \ a{21} & a{22} & a{23} \ a{31} & a{32} & a{33} \end{bmatrix} \rightarrow \text{After Gaussian elimination} \rightarrow A = \begin{bmatrix} 1 & b{12} & b{13} \ 0 & 1 & b{23} \ 0 & 0 & 1 \end{bmatrix}

  • First step: Obtain a 1 as the first entry.

How To: Perform Row Operations to Achieve Row-Echelon Form

  1. First equation should have a leading coefficient of 1 (interchange rows or multiply by a constant if needed).

  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.

  7. Place any all-zero rows at the bottom.

Example: Performing Row Operations on a 3x3 Augmented Matrix
  • Given matrix:
    [1amp;3amp;4amp;3 3amp;2amp;5amp;6 3amp;3amp;4amp;6]\begin{bmatrix} 1 & -3 & 4 & 3 \ 3 & 2 & -5 & 6 \ -3 & 3 & 4 & 6 \end{bmatrix}

  • Step 1: The first row already has a 1 in row 1, column 1.

  • Step 2: Multiply row 1 by -2 and add it to row 2, replace row 2 with the result:
    2R1+R2=R2[1amp;3amp;4amp;3 0amp;8amp;13amp;0 3amp;3amp;4amp;6]-2R1 + R2 = R_2 \rightarrow \begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 8 & -13 & 0 \ -3 & 3 & 4 & 6 \end{bmatrix}

  • Step 3: Obtain a zero in row 3, column 1:
    3R1+R3=R3[1amp;3amp;4amp;3 0amp;11amp;17amp;15 0amp;6amp;16amp;15]3R1 + R3 = R_3 \rightarrow \begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 11 & -17 & 15 \ 0 & -6 & 16 & 15 \end{bmatrix}
    [1amp;3amp;4amp;3 0amp;1amp;2amp;0 0amp;3amp;3amp;4amp;6]\begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 1 & -2 & 0 \ 0 & -3 & 3 & 4 & 6 \end{bmatrix}

  • Step 4: Obtain a zero in row 3, column 2:
    6R2+R3=R3[1amp;3amp;4amp;3 0amp;1amp;2amp;0 0amp;0amp;4amp;15]6R2 + R3 = R_3 \rightarrow \begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 1 & -2 & 0 \ 0 & 0 & 4 & 15 \end{bmatrix}

  • Step 5: Obtain a 1 in row 3, column 3:
    14R3=R3[1amp;3amp;4amp;3 0amp;1amp;2amp;0 0amp;0amp;1amp;154]\frac{1}{4}R3 = R3 \rightarrow \begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 1 & -2 & 0 \ 0 & 0 & 1 & \frac{15}{4} \end{bmatrix}

Try It
  • System of equations in row-echelon form:
    x2y+3z=9x - 2y + 3z = 9
    x+3y=4-x + 3y = -4
    2x5y+5z=172x - 5y + 5z = 17