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:3 x + 4 y = 7 3x + 4y = 7 3 x + 4 y = 7 4 x − 2 y = 5 4x - 2y = 5 4 x − 2 y = 5
Augmented matrix:[ 3 a m p ; 4 a m p ; 7 4 a m p ; − 2 a m p ; 5 ] \begin{bmatrix} 3 & 4 & 7 \ 4 & -2 & 5 \end{bmatrix} [ 3 am p ; 4 am p ; 7 4 am p ; − 2 am p ; 5 ]
Coefficient Matrix Example: 3x3 System System of equations:3 x − y − z = 0 3x - y - z = 0 3 x − y − z = 0 x + y = 5 x + y = 5 x + y = 5 2 x − 3 z = 2 2x - 3z = 2 2 x − 3 z = 2
Coefficient matrix:
[ 3 a m p ; − 1 a m p ; − 1 1 a m p ; 1 a m p ; 0 2 a m p ; 0 a m p ; − 3 ] \begin{bmatrix} 3 & -1 & -1 \\ 1 & 1 & 0 \\ 2 & 0 & -3 \end{bmatrix} 3 1 2 am p ; − 1 am p ; 1 am p ; 0 am p ; − 1 am p ; 0 am p ; − 3
Augmented matrix:[ 3 a m p ; − 1 a m p ; − 1 a m p ; 0 1 a m p ; 1 a m p ; 0 a m p ; 5 2 a m p ; 0 a m p ; − 3 a m p ; 2 ] \begin{bmatrix} 3 & -1 & -1 & 0 \ 1 & 1 & 0 & 5 \ 2 & 0 & -3 & 2 \end{bmatrix} [ 3 am p ; − 1 am p ; − 1 am p ; 0 1 am p ; 1 am p ; 0 am p ; 5 2 am p ; 0 am p ; − 3 am p ; 2 ] y, z terms).
Equations must be in standard form: a x + b y + c z = d ax + by + cz = d a x + b y + cz = d
Missing variable terms have a coefficient of 0.
How To: Write an Augmented Matrix Write coefficients of x-terms down the first column.
Write coefficients of y-terms down the second column.
If there are z-terms, write coefficients down the third column.
Draw a vertical line and write the constants to the right.
Example: Writing the Augmented Matrix System of equations:x + 2 y − z = 3 x + 2y - z = 3 x + 2 y − z = 3 2 x − y + 2 z = 6 2x - y + 2z = 6 2 x − y + 2 z = 6 x − 3 y + 3 z = 4 x - 3y + 3z = 4 x − 3 y + 3 z = 4
Augmented matrix:[ 1 a m p ; 2 a m p ; − 1 a m p ; 3 2 a m p ; − 1 a m p ; 2 a m p ; 6 1 a m p ; − 3 a m p ; 3 a m p ; 4 ] \begin{bmatrix} 1 & 2 & -1 & 3 \ 2 & -1 & 2 & 6 \ 1 & -3 & 3 & 4 \end{bmatrix} [ 1 am p ; 2 am p ; − 1 am p ; 3 2 am p ; − 1 am p ; 2 am p ; 6 1 am p ; − 3 am p ; 3 am p ; 4 ]
Try It Writing a System of Equations from an Augmented Matrix Example: Writing a System of Equations Augmented matrix:[ 1 a m p ; − 3 a m p ; − 5 a m p ; − 2 2 a m p ; − 5 a m p ; − 4 a m p ; 5 − 3 a m p ; 5 a m p ; 4 a m p ; 6 ] \begin{bmatrix} 1 & -3 & -5 & -2 \ 2 & -5 & -4 & 5 \ -3 & 5 & 4 & 6 \end{bmatrix} [ 1 am p ; − 3 am p ; − 5 am p ; − 2 2 am p ; − 5 am p ; − 4 am p ; 5 − 3 am p ; 5 am p ; 4 am p ; 6 ]
System of equations:x − 3 y − 5 z = − 2 x - 3y - 5z = -2 x − 3 y − 5 z = − 2 2 x − 5 y − 4 z = 5 2x - 5y - 4z = 5 2 x − 5 y − 4 z = 5 − 3 x + 5 y + 4 z = 6 -3x + 5y + 4z = 6 − 3 x + 5 y + 4 z = 6
Try It 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:[ 1 a m p ; a a m p ; b 0 a m p ; 1 a m p ; d 0 a m p ; 0 a m p ; 1 ] \begin{bmatrix} 1 & a & b \ 0 & 1 & d \ 0 & 0 & 1 \end{bmatrix} [ 1 am p ; a am p ; b 0 am p ; 1 am p ; d 0 am p ; 0 am p ; 1 ]
Guidelines for Obtaining Row-Echelon Form First nonzero number in any nonzero row is a 1 (leading 1).
All-zero rows are at the bottom.
Any leading 1 is below and to the right of a previous leading 1.
Any column containing a leading 1 has zeros in all other positions.
Row Operations Notation Interchange rows: R i ↔ R j Ri \leftrightarrow Rj R i ↔ R j
Multiply a row by a constant: c R i cR_i c R i
Add the product of a row multiplied by a constant to another row: R i + c R j Ri + cRj R i + c R j
Gaussian Elimination Method to obtain row-echelon form.
Goal: Write matrix A with 1s down the main diagonal and zeros below.
A = [ a 11 a m p ; a 12 a m p ; a 13 a 21 a m p ; a 22 a m p ; a 23 a 31 a m p ; a 32 a m p ; a 33 ] → After Gaussian elimination → A = [ 1 a m p ; b 12 a m p ; b 13 0 a m p ; 1 a m p ; b 23 0 a m p ; 0 a m p ; 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} A = [ a 11 am p ; a 12 am p ; a 13 a 21 am p ; a 22 am p ; a 23 a 31 am p ; a 32 am p ; a 33 ] → After Gaussian elimination → A = [ 1 am p ; b 12 am p ; b 13 0 am p ; 1 am p ; b 23 0 am p ; 0 am p ; 1 ]
First step: Obtain a 1 as the first entry.
How To: Perform Row Operations to Achieve Row-Echelon Form First equation should have a leading coefficient of 1 (interchange rows or multiply by a constant if needed).
Use row operations to obtain zeros down the first column below the first entry of 1.
Use row operations to obtain a 1 in row 2, column 2.
Use row operations to obtain zeros down column 2, below the entry of 1.
Use row operations to obtain a 1 in row 3, column 3.
Continue this process for all rows.
Place any all-zero rows at the bottom.
Example: Performing Row Operations on a 3x3 Augmented Matrix Given matrix:[ 1 a m p ; − 3 a m p ; 4 a m p ; 3 3 a m p ; 2 a m p ; − 5 a m p ; 6 − 3 a m p ; 3 a m p ; 4 a m p ; 6 ] \begin{bmatrix} 1 & -3 & 4 & 3 \ 3 & 2 & -5 & 6 \ -3 & 3 & 4 & 6 \end{bmatrix} [ 1 am p ; − 3 am p ; 4 am p ; 3 3 am p ; 2 am p ; − 5 am p ; 6 − 3 am p ; 3 am p ; 4 am p ; 6 ]
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:− 2 R 1 + R 2 = R 2 → [ 1 a m p ; − 3 a m p ; 4 a m p ; 3 0 a m p ; 8 a m p ; − 13 a m p ; 0 − 3 a m p ; 3 a m p ; 4 a m p ; 6 ] -2R1 + R2 = R_2 \rightarrow \begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 8 & -13 & 0 \ -3 & 3 & 4 & 6 \end{bmatrix} − 2 R 1 + R 2 = R 2 → [ 1 am p ; − 3 am p ; 4 am p ; 3 0 am p ; 8 am p ; − 13 am p ; 0 − 3 am p ; 3 am p ; 4 am p ; 6 ]
Step 3: Obtain a zero in row 3, column 1:3 R 1 + R 3 = R 3 → [ 1 a m p ; − 3 a m p ; 4 a m p ; 3 0 a m p ; 11 a m p ; − 17 a m p ; 15 0 a m p ; − 6 a m p ; 16 a m p ; 15 ] 3R1 + R3 = R_3 \rightarrow \begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 11 & -17 & 15 \ 0 & -6 & 16 & 15 \end{bmatrix} 3 R 1 + R 3 = R 3 → [ 1 am p ; − 3 am p ; 4 am p ; 3 0 am p ; 11 am p ; − 17 am p ; 15 0 am p ; − 6 am p ; 16 am p ; 15 ] [ 1 a m p ; − 3 a m p ; 4 a m p ; 3 0 a m p ; 1 a m p ; − 2 a m p ; 0 0 a m p ; − 3 a m p ; 3 a m p ; 4 a m p ; 6 ] \begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 1 & -2 & 0 \ 0 & -3 & 3 & 4 & 6 \end{bmatrix} [ 1 am p ; − 3 am p ; 4 am p ; 3 0 am p ; 1 am p ; − 2 am p ; 0 0 am p ; − 3 am p ; 3 am p ; 4 am p ; 6 ]
Step 4: Obtain a zero in row 3, column 2:6 R 2 + R 3 = R 3 → [ 1 a m p ; − 3 a m p ; 4 a m p ; 3 0 a m p ; 1 a m p ; − 2 a m p ; 0 0 a m p ; 0 a m p ; 4 a m p ; 15 ] 6R2 + R3 = R_3 \rightarrow \begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 1 & -2 & 0 \ 0 & 0 & 4 & 15 \end{bmatrix} 6 R 2 + R 3 = R 3 → [ 1 am p ; − 3 am p ; 4 am p ; 3 0 am p ; 1 am p ; − 2 am p ; 0 0 am p ; 0 am p ; 4 am p ; 15 ]
Step 5: Obtain a 1 in row 3, column 3:1 4 R 3 = R 3 → [ 1 a m p ; − 3 a m p ; 4 a m p ; 3 0 a m p ; 1 a m p ; − 2 a m p ; 0 0 a m p ; 0 a m p ; 1 a m p ; 15 4 ] \frac{1}{4}R3 = R3 \rightarrow \begin{bmatrix} 1 & -3 & 4 & 3 \ 0 & 1 & -2 & 0 \ 0 & 0 & 1 & \frac{15}{4} \end{bmatrix} 4 1 R 3 = R 3 → [ 1 am p ; − 3 am p ; 4 am p ; 3 0 am p ; 1 am p ; − 2 am p ; 0 0 am p ; 0 am p ; 1 am p ; 4 15 ]
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