A system of equations can be solved by using Gauss-Jordan Elimination. This process involves applying row operations to achieve the Reduced Row-Echelon Form
First, write the system of equations in matrix form by taking the constants
x+2y+z=9
3y+z=10
x+y+2z=11
[121|9031|10112|11]R1R2R3
Use row operations for the matrix to achieve Row-Echelon Form
Start with transforming the first column
Transforming the First Element of R3:
[121|9031|10112|11]R1R2R3
R1−R3→R3
R1:121|9−R3:−1−1−2|−11R3→01−1|−2
New matrix:
[121|9031|1001−1|−2]R1R2R3
Next, transform R2
Transforming the Third Element of R2:
[121|9031|1001−1|−2]R1R2R3
R3+R2→R2
R3:01−1|−2R2:031|10R2→040|8
Transforming the Second Element of R2:
[121|9040|801−1|−2]R1R2R3
R2÷4→R2
R2÷4→010|2
New matrix:
[121|9010|201−1|−2]R1R2R3
Proceed with transforming R3
Transforming the Second Element of R2:
[121|9010|201−1|−2]R1R2R3
−1R2+R3→R3
−1R2:0−10|−2R3:01−1|−2R3→00−1|−4
Transforming the Third Element of R3:
[121|9010|200−1|−4]R1R2R3
R3÷−1→R3
R3÷−1→001|4
New matrix:
[121|9010|2001|4]R1R2R3
Now, there is one row left to transform which is R1
Transforming the Third Element of R1:
[121|9010|2001|4]R1R2R3
−1R3+R1→R1
−1R3:00−1|−4R1:121|9R1→120|5
Transforming the Second Element of R1:
[120|5010|2001|4]R1R2R3
−2R2+R1→R1
−2R2:0−20|−4R1:120|5R1→100|1
New matrix:
[100|1010|2001|4]R1R2R3
This new matrix is in Reduced Row-Echelon Form
Remember that each row in this matrix corresponds to a term in an equation
Convert the rows into equations
xyzC [100|1010|2001|4]
x=1y=2z=4
x=1
y=2
z=4
Question 2 of 2
2. Question
Solve for x,y and z
x-3y-z=-9
2y+z=2
-x-y=-1
x=(-3)
y=(4)
z=(-6)
Hint
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Reduced Row-Echelon Form
[100x010x001x]
A system of equations can be solved by using Gauss-Jordan Elimination. This process involves applying row operations to achieve the Reduced Row-Echelon Form
First, write the system of equations in matrix form by taking the constants
x-3y-z=-9
2y+z=2
-x-y=-1
[1−3−1|−9021|2−1−10|−1]R1R2R3
Use row operations for the matrix to achieve Row-Echelon Form
Start with swapping R1 and R3 since R3 already has 0 as its third element
[−1−10|−1021|21−3−1|−9]R1R2R3
Next, transform R3
Transforming the First Element of R3:
[−1−10|−1021|21−3−1|−9]R1R2R3
−R1−R3→R3
−R1:110|1−R3:−131|9R3→041|10
Transforming the Second Element of R3:
[−1−10|−1021|2041|10]R1R2R3
2R2−R3→R3
2R2:042|4−R3:0−4−1|−10R3→001|−6
New matrix:
[−1−10|−1021|2001|−6]R1R2R3
Proceed with transforming R2
Transforming the Third Element of R2:
[−1−10|−1021|2001|−6]R1R2R3
−R3+R2→R2
−R3:00−1|6R2:021|2R2→020|8
Transforming the Second Element of R2:
[−1−10|−1020|8001|−6]R1R2R3
R2÷2→R2
R2÷2→010|4
New matrix:
[−1−10|−1010|4001|−6]R1R2R3
Now, there is one row left to transform which is R1
Transforming the Second Element of R1:
[−1−10|−1010|4001|−6]R1R2R3
R2+R1→R1
R2:010|4R1:−1−10|−1R1→−100|3
Transforming the First Element of R1:
[−100|3010|4001|−6]R1R2R3
R1÷−1→R2
R2÷2→100|−3
New matrix:
[100|−3010|4001|−6]R1R2R3
This new matrix is in Reduced Row-Echelon Form
Remember that each row in this matrix corresponds to a term in an equation