Math 1313
Section 3.2
Example 6:Solve the system of linear equations using the GaussJordan elimination method. 12z5y73z3y2x9z8y==+=Popper 2: Is the following matrix in row reduced form?
°
0
1
0
0
3
0
0
1
0
3
0
0
0
1
7
0
0
0
0
0
±
Math 1313
Section 3.2
Infinite Number of Solutions
Example 7:The following augmented matrix in rowreduced form is equivalent to the augmented matrix of a certain system of linear equations. Use this result to solve the system of equations. 000025103101Example 8:Solve the system of linear equations using the GaussJordan elimination method.
2
Math 1313
Section 3.2
A System of Equations That Has No Solution
In using the GaussJordan elimination method the following equivalent matrix was obtained (note
this matrix is not in rowreduced form, let’s see why):



1
0
0
0
1
4
4
0
1
1
1
1
Look at the last row.
It reads:
0x + 0y + 0z = 1, in other words, 0 = 1!!!
This is never true.
So
the system is inconsistent and has no solution.
Mark D for question 4
Systems with No Solution
If there is a row in the augmented matrix containing all zeros to the left of the vertical line and a nonzero entry to the right of the line, then the system of equations has no solution.
Example 9: Solve the system of linear equations using the GaussJordan elimination method.
32323==+=yxyxyx
2
Math 1313
Section 3.2
Example 10:Solve the system of linear equations using the GaussJordan elimination method.
−² + 3³ − 4´ = 124² − 12³ + 16´ = −36
Example 11: Solve the system of linear equations using the GaussJordan elimination method.
2² − 3³ = 13² + ³ = −1² − 4³ = 14Popper 3
: State the number of solutions, if any.
µ
1
3
1
¶
a.One Solution b.Infinitely Many Solutions c.No Solution
Math 1313
Section 3.2
Popper 5: State the number of solutions, if any. 2² − ³ + 3´ = 4−6² + 3³ − 9³ = −12
a.One Solution b.Infinitely Many Solutions c.No Solution