4.3 Gauss Elimination for Systems of Linear Equations
MATH 166-506, Fall 2016, Jean Yeh
4.3 Gauss Elimination for Systems of Linear Equations
In this section, we are going to briefly introduce the idea of system of linear equation and the Gauss
elimination. (Check your textbook for more detail)
The linear equation of 3 variables is form as below:
( for example: 2x + 5y − 8z = 1)
ax + by + cz = d,
A system of linear equations is a system contains more than one linear equations. To solve the
system of linear equations is to find a set of points satisfies ALL the equations in the system.
We can use algebraic methods to solve systems of equations, such as substitution and elimination.
We can also introduce the idea of MATRIX and use the Gauss Elimination method to solve it.
The matrix is a rectangular array of numbers. A matrix conatins n rows and m columns if the
size of the matrix is n × m. We can represent a system of equations with an augmented matrix.
An augmented matrix is a short-hand way of representing a system without having to write the
variables.
x + 2y − 2z =
1
2z + 7y + 2z = −1
x + 6y + 7z = −3

⇒

1 2 −2
1
 2 7
2 −1 
1 6
7 −3
Example: Solve the following system of equation.
x + 2y = 14......(1)
2x + 3y = 25......(2)
ANSWER (A)
by (1), we have x = 14 − 2y......(1*), substitute back into (2) and obtain.
2(14 − 2y) + 3y = 25 ⇒ 28 − 4y + 3y = 25 ⇒ y = 3
Now subtitue into (1*) and obtain
x + 2(3) = 14 ⇒ x = 8
ANSWER
(B)
E1 : x+ 2y = 14
E2 : 2x+ 3y = 25
E3 =2×E1 −E2
========⇒
E1 : x+ 2y = 14
E3 :
y = 3
E4 =E1 −2×E2
========⇒
E4 : x
= 8
E3 :
y = 3
ANSWER
(C)
1 2 14
1 2 14
1 0 8
2R1 −R2 →R2
R1 −2R2 →R1
=======⇒
=======⇒
2 3 25
0 1 3
0 1 3
Page 1
4.3 Gauss Elimination for Systems of Linear Equations
MATH 166-506, Fall 2016, Jean Yeh
Using Calculator to solve Matrix by rref
To enter a matrix into your calculator and find the row-reduced form:
Press 2ND and x−1 to select MATRIX
Cursor over to EDIT and select where( A , B, etc) you want to store your matrix.
Enter the size of the matrix and then enter the value in the matrix.
Press 2ND and MODE to return to your home screen.
Press 2ND and x−1 to select MATRIX again.
Cursor right to MATH .
Scroll down and select rref (option B).
6. Press 2ND and x−1 to select MATRIX one more time.
Under the NAMES column, select the matrix you want to reduce.
7. Close your parentheses and press ENTER .
1.
2.
3.
4.
5.
Example: Use calculator to solve the following system of equation.
x + 2y = 14
2x + 3y = 25
rref(
1 2 14
2 3 25
)=
1 0 8
0 1 3
How to Clear Matrices
1. Press the 2nd and the + .
2. Scroll to Mem Mgmt/Del .
3. Press the ENTER .
4. Select Matrix (option 5) and press the ENTER .
5. Scroll to each matrix and press DEL . This will clear the matrix out of the memory.
There’s 3 possibilities for the number of solutions for a system of linear equations:
• Exactly One Solution (called a unique solution).
• No Solution (This is called an inconsistent system.)
• Infinitely Many Solutions (This is called a dependent system.)(will discuss in section 4.4.)
Example: Represent the systems of equations with an augmented matrix and then solve it.
3x + 2y − 6z = 14
4x − 5y + 2z = 3
7x + 9y + 3z = 14




3
2 −6 14
1 0 0
1.9309
2 3 )= 0 1 0
0.4587 
rref(  4 −5
7
9
3 14
0 0 1 −1.2150
x = 1.9309
Therefore, y = 0.4587
z = −1.2150
Page 2
4.3 Gauss Elimination for Systems of Linear Equations
MATH 166-506, Fall 2016, Jean Yeh
Example: Represent the systems of equations with an augmented matrix and then solve it.
2x − 4y = 9
−2y − z = 0
x + y + 4 = 2z
3y + z = 5x
Rewrite the system as
2x


rref( 

2
4
0
0 −2 −1
1
1 −2
−5
3
1
4y
=
9
− 2y − z =
0
x + y − 2z = −4
−5x + 3y + z =
0



9
1 0 0 0


0 
)= 0 1 0 0 
Therefore, there’s NO solution.
 0 0 1 0 
−4 
0
0 0 0 1
Page 3