2.5B Augmented matrices and Reduced Row-Echelon Form
Objectives:
1. Find reduced row-echelon form of an augmented matrix to solve
systems of equations.
2.5B Augmented matrices and Reduced Row-Echelon Form
An augmented matrix is composed of columns representing the
coefficients of each variable and the constant term.
Write the system of equations as an augmented matrix.
x  2y  z  7
6x  2 y  2z  4
4 x  6 y  4 z  14
1  2 1 7 
6 2  2 4 


4 6
4 14
2.5B Augmented matrices and Reduced Row-Echelon Form
We will use reduced row-echelon form to solve this augmented matrix.
The goal is to get the identity matrix in the first three columns (to the
left of the line) and the solution will be to the right of the line.
1 0 0
0 1 0

0 0 1
x
y 
z 
We do this by adding and subtracting rows or multiples of rows from
each other. This will give us zeros in every spot but along the diagonal.
2.5B Augmented matrices and Reduced Row-Echelon Form
1  2 1 7 
6 2  2 4 


4 6
4 14
For a new row 2: Add -6 times rows 1 and row 2 together
(-6R1 + R2)
For a new row 3: Add -4 times rows 1 and
row 3 together (-4R1 + R3)
7 
1  2 1
0 14  8  38


0 14 0  14 
7 
1  2 1
0 14  8  38


4 6
4
14 
For a new row 3: multiple row 3 by 1/14
(1/14R3)
7 
1  2 1
0 14  8  38


0 1
0
 1 
2.5B Augmented matrices and Reduced Row-Echelon Form
7 
1  2 1
0 14  8  38


0 1
0
 1 
For a new row 1: Add 2 times row 3 and row 1 together
(2R3 + R1)
For a new row 2: Add -13 times rows 3 and row 2
together (-13R3 + R2)
For a new row 2 and row 3: Switch row 2
and row 3
5 
1 0 1
0 1 0

1 

0 1  8  25
5 
1 0 1
0 1  8  25


0 1 0
 1 
For a new row 3: Subtract row 3 from row 2
(R2 – R3)
1 0 1 5 
0 1 0  1


0 0 8 24 
2.5B Augmented matrices and Reduced Row-Echelon Form
1 0 1 5 
0 1 0  1


0 0 8 24 
For a new row 3: Multiple row 3 by 1/8
(1/8R3)
For a new row 1: Subtract row 3 from row 1
(R1 – R3)
1 0 0 2 
0 1 0  1


0 0 1 3 
1 0 1 5 
0 1 0  1


0 0 1 3 
Now we have the augmented matrix in
reduced row-echelon form. And we can see
the solution.
x=2
y = -1
z = 3 or (2, -1, 3)
Check the solutions with the system of equations.
2.5B Augmented matrices and Reduced Row-Echelon Form
Now let’s use the graphing calculator to check answers.
Enter each equation as a row of the matrix A:
x  2y  z  7
6x  2 y  2z  4
4 x  6 y  4 z  14
Hit 2nd MATRX go to EDIT Select 1 for [A]. Enter the dimensions for the
augmented matrix which is 3 x 4 (Row x Column). Then enter all the
coefficients and the constants. Once you are done go to 2nd MATRX go
to MATH then find rref (which stands for reduced row-echelon form).
The screen should have rref( now enter 2nd MATRX go to NAMES Select
1 for [A] then close the parenthesis and press enter.
NOTE: Using the calculator does not mean you do not do the work. You use the calculator to double check
answers. The answer is not as important as the process.
Date
Assignment number
106
1-5
5
Read pages 107-109 for tomorrow.