math 1300: section 4- 3 gauss-jordan elimination
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![Page 1: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/1.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Math 1300 Finite MathematicsSection 4.3 Gauss-Jordan Elimination
Jason Aubrey
Department of MathematicsUniversity of Missouri
Jason Aubrey Math 1300 Finite Mathematics
![Page 2: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/2.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
In the previous section, we used row operations to transformthe augmented coefficient matrix for a system of two equationsin two variables
[a11 a12a21 a22
∣∣∣∣k1k2
]a11x1 + a12x2 = k1
a21x1 + a22x2 = k2
into one of the following simplified forms:[1 00 1
∣∣∣∣mn] [
1 m0 0
∣∣∣∣n0] [
1 m0 0
∣∣∣∣np]
p 6= 0
We can classify the solutions based on the three simplifiedforms.
Jason Aubrey Math 1300 Finite Mathematics
![Page 3: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/3.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
In the previous section, we used row operations to transformthe augmented coefficient matrix for a system of two equationsin two variables
[a11 a12a21 a22
∣∣∣∣k1k2
]a11x1 + a12x2 = k1
a21x1 + a22x2 = k2
into one of the following simplified forms:[1 00 1
∣∣∣∣mn] [
1 m0 0
∣∣∣∣n0] [
1 m0 0
∣∣∣∣np]
p 6= 0
We can classify the solutions based on the three simplifiedforms.
Jason Aubrey Math 1300 Finite Mathematics
![Page 4: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/4.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
In the previous section, we used row operations to transformthe augmented coefficient matrix for a system of two equationsin two variables
[a11 a12a21 a22
∣∣∣∣k1k2
]a11x1 + a12x2 = k1
a21x1 + a22x2 = k2
into one of the following simplified forms:[1 00 1
∣∣∣∣mn] [
1 m0 0
∣∣∣∣n0] [
1 m0 0
∣∣∣∣np]
p 6= 0
We can classify the solutions based on the three simplifiedforms.
Jason Aubrey Math 1300 Finite Mathematics
![Page 5: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/5.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Summary
Form 1: A Unique Solution (Consistent and Independent)[1 00 1
∣∣∣∣mn]
Form 2: Infinitely Many Solutons (Consistent and Dependent)[1 m0 0
∣∣∣∣n0]
Form 3: No Solution (Inconsistent)[1 m0 0
∣∣∣∣np]
p 6= 0
Jason Aubrey Math 1300 Finite Mathematics
![Page 6: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/6.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Summary
Form 1: A Unique Solution (Consistent and Independent)[1 00 1
∣∣∣∣mn]
Form 2: Infinitely Many Solutons (Consistent and Dependent)[1 m0 0
∣∣∣∣n0]
Form 3: No Solution (Inconsistent)[1 m0 0
∣∣∣∣np]
p 6= 0
Jason Aubrey Math 1300 Finite Mathematics
![Page 7: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/7.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Summary
Form 1: A Unique Solution (Consistent and Independent)[1 00 1
∣∣∣∣mn]
Form 2: Infinitely Many Solutons (Consistent and Dependent)[1 m0 0
∣∣∣∣n0]
Form 3: No Solution (Inconsistent)[1 m0 0
∣∣∣∣np]
p 6= 0
Jason Aubrey Math 1300 Finite Mathematics
![Page 8: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/8.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
DefinitionA matrix is said to be in reduced row echelon form, or, moresimply, in reduced form, if
Each row consisting entirely of zeros is below any rowhaving at least one nonzero element.The leftmost nonzero element in each row is 1.All other elements in the column containing the leftmost 1of a given row are zeros.The leftmost 1 in any row is to the right of the leftmost 1 inthe row above.
Jason Aubrey Math 1300 Finite Mathematics
![Page 9: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/9.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
DefinitionA matrix is said to be in reduced row echelon form, or, moresimply, in reduced form, if
Each row consisting entirely of zeros is below any rowhaving at least one nonzero element.
The leftmost nonzero element in each row is 1.All other elements in the column containing the leftmost 1of a given row are zeros.The leftmost 1 in any row is to the right of the leftmost 1 inthe row above.
Jason Aubrey Math 1300 Finite Mathematics
![Page 10: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/10.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
DefinitionA matrix is said to be in reduced row echelon form, or, moresimply, in reduced form, if
Each row consisting entirely of zeros is below any rowhaving at least one nonzero element.The leftmost nonzero element in each row is 1.
All other elements in the column containing the leftmost 1of a given row are zeros.The leftmost 1 in any row is to the right of the leftmost 1 inthe row above.
Jason Aubrey Math 1300 Finite Mathematics
![Page 11: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/11.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
DefinitionA matrix is said to be in reduced row echelon form, or, moresimply, in reduced form, if
Each row consisting entirely of zeros is below any rowhaving at least one nonzero element.The leftmost nonzero element in each row is 1.All other elements in the column containing the leftmost 1of a given row are zeros.
The leftmost 1 in any row is to the right of the leftmost 1 inthe row above.
Jason Aubrey Math 1300 Finite Mathematics
![Page 12: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/12.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
DefinitionA matrix is said to be in reduced row echelon form, or, moresimply, in reduced form, if
Each row consisting entirely of zeros is below any rowhaving at least one nonzero element.The leftmost nonzero element in each row is 1.All other elements in the column containing the leftmost 1of a given row are zeros.The leftmost 1 in any row is to the right of the leftmost 1 inthe row above.
Jason Aubrey Math 1300 Finite Mathematics
![Page 13: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/13.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Determine if each of the following matrices is inreduced row echelon form.
(a)[1 00 1
∣∣∣∣ 2−3
]is in reduced form.
(b)[1 20 1
∣∣∣∣ 2−3
]is not in reduced form.
(c)
1 0 00 1 00 0 1
∣∣∣∣∣∣2−13
is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 14: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/14.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Determine if each of the following matrices is inreduced row echelon form.
(a)[1 00 1
∣∣∣∣ 2−3
]
is in reduced form.
(b)[1 20 1
∣∣∣∣ 2−3
]is not in reduced form.
(c)
1 0 00 1 00 0 1
∣∣∣∣∣∣2−13
is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 15: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/15.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Determine if each of the following matrices is inreduced row echelon form.
(a)[1 00 1
∣∣∣∣ 2−3
]is in reduced form.
(b)[1 20 1
∣∣∣∣ 2−3
]is not in reduced form.
(c)
1 0 00 1 00 0 1
∣∣∣∣∣∣2−13
is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 16: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/16.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Determine if each of the following matrices is inreduced row echelon form.
(a)[1 00 1
∣∣∣∣ 2−3
]is in reduced form.
(b)[1 20 1
∣∣∣∣ 2−3
]
is not in reduced form.
(c)
1 0 00 1 00 0 1
∣∣∣∣∣∣2−13
is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 17: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/17.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Determine if each of the following matrices is inreduced row echelon form.
(a)[1 00 1
∣∣∣∣ 2−3
]is in reduced form.
(b)[1 20 1
∣∣∣∣ 2−3
]is not in reduced form.
(c)
1 0 00 1 00 0 1
∣∣∣∣∣∣2−13
is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 18: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/18.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Determine if each of the following matrices is inreduced row echelon form.
(a)[1 00 1
∣∣∣∣ 2−3
]is in reduced form.
(b)[1 20 1
∣∣∣∣ 2−3
]is not in reduced form.
(c)
1 0 00 1 00 0 1
∣∣∣∣∣∣2−13
is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 19: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/19.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Determine if each of the following matrices is inreduced row echelon form.
(a)[1 00 1
∣∣∣∣ 2−3
]is in reduced form.
(b)[1 20 1
∣∣∣∣ 2−3
]is not in reduced form.
(c)
1 0 00 1 00 0 1
∣∣∣∣∣∣2−13
is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 20: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/20.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
(d)
1 0 00 0 00 0 1
∣∣∣∣∣∣203
is not in reduced form.
(e)
0 1 00 0 30 0 0
∣∣∣∣∣∣2−10
is not in reduced form.
(f)
1 4 −10 0 10 0 0
∣∣∣∣∣∣310
is not in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 21: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/21.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
(d)
1 0 00 0 00 0 1
∣∣∣∣∣∣203
is not in reduced form.
(e)
0 1 00 0 30 0 0
∣∣∣∣∣∣2−10
is not in reduced form.
(f)
1 4 −10 0 10 0 0
∣∣∣∣∣∣310
is not in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 22: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/22.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
(d)
1 0 00 0 00 0 1
∣∣∣∣∣∣203
is not in reduced form.
(e)
0 1 00 0 30 0 0
∣∣∣∣∣∣2−10
is not in reduced form.
(f)
1 4 −10 0 10 0 0
∣∣∣∣∣∣310
is not in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 23: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/23.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
(d)
1 0 00 0 00 0 1
∣∣∣∣∣∣203
is not in reduced form.
(e)
0 1 00 0 30 0 0
∣∣∣∣∣∣2−10
is not in reduced form.
(f)
1 4 −10 0 10 0 0
∣∣∣∣∣∣310
is not in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 24: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/24.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
(d)
1 0 00 0 00 0 1
∣∣∣∣∣∣203
is not in reduced form.
(e)
0 1 00 0 30 0 0
∣∣∣∣∣∣2−10
is not in reduced form.
(f)
1 4 −10 0 10 0 0
∣∣∣∣∣∣310
is not in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 25: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/25.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
(d)
1 0 00 0 00 0 1
∣∣∣∣∣∣203
is not in reduced form.
(e)
0 1 00 0 30 0 0
∣∣∣∣∣∣2−10
is not in reduced form.
(f)
1 4 −10 0 10 0 0
∣∣∣∣∣∣310
is not in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 26: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/26.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination
3x1 + x2 − 2x3 = 2x1 − 2x2 + x3 = 3
2x1 − x2 − 3x3 = 3
3 1 −21 −2 12 −1 −3
∣∣∣∣∣∣233
R1↔R2−−−−→
1 −2 13 1 −22 −1 −3
∣∣∣∣∣∣323
Jason Aubrey Math 1300 Finite Mathematics
![Page 27: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/27.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination
3x1 + x2 − 2x3 = 2x1 − 2x2 + x3 = 3
2x1 − x2 − 3x3 = 3
3 1 −21 −2 12 −1 −3
∣∣∣∣∣∣233
R1↔R2−−−−→
1 −2 13 1 −22 −1 −3
∣∣∣∣∣∣323
Jason Aubrey Math 1300 Finite Mathematics
![Page 28: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/28.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination
3x1 + x2 − 2x3 = 2x1 − 2x2 + x3 = 3
2x1 − x2 − 3x3 = 3
3 1 −21 −2 12 −1 −3
∣∣∣∣∣∣233
R1↔R2−−−−→
1 −2 13 1 −22 −1 −3
∣∣∣∣∣∣323
Jason Aubrey Math 1300 Finite Mathematics
![Page 29: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/29.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination
3x1 + x2 − 2x3 = 2x1 − 2x2 + x3 = 3
2x1 − x2 − 3x3 = 3
3 1 −21 −2 12 −1 −3
∣∣∣∣∣∣233
R1↔R2−−−−→
1 −2 13 1 −22 −1 −3
∣∣∣∣∣∣323
Jason Aubrey Math 1300 Finite Mathematics
![Page 30: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/30.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
−3R1+R2→R2−−−−−−−−−→
1 −2 10 7 −52 −1 −3
∣∣∣∣∣∣3−73
−2R1+R3→R3−−−−−−−−−→
1 −2 10 7 −50 3 −5
∣∣∣∣∣∣3−7−3
−2R3+R2→R2−−−−−−−−−→
1 −2 10 1 50 3 −5
∣∣∣∣∣∣3−1−3
Jason Aubrey Math 1300 Finite Mathematics
![Page 31: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/31.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
−3R1+R2→R2−−−−−−−−−→
1 −2 10 7 −52 −1 −3
∣∣∣∣∣∣3−73
−2R1+R3→R3−−−−−−−−−→
1 −2 10 7 −50 3 −5
∣∣∣∣∣∣3−7−3
−2R3+R2→R2−−−−−−−−−→
1 −2 10 1 50 3 −5
∣∣∣∣∣∣3−1−3
Jason Aubrey Math 1300 Finite Mathematics
![Page 32: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/32.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
−3R1+R2→R2−−−−−−−−−→
1 −2 10 7 −52 −1 −3
∣∣∣∣∣∣3−73
−2R1+R3→R3−−−−−−−−−→
1 −2 10 7 −50 3 −5
∣∣∣∣∣∣3−7−3
−2R3+R2→R2−−−−−−−−−→
1 −2 10 1 50 3 −5
∣∣∣∣∣∣3−1−3
Jason Aubrey Math 1300 Finite Mathematics
![Page 33: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/33.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
−3R1+R2→R2−−−−−−−−−→
1 −2 10 7 −52 −1 −3
∣∣∣∣∣∣3−73
−2R1+R3→R3−−−−−−−−−→
1 −2 10 7 −50 3 −5
∣∣∣∣∣∣3−7−3
−2R3+R2→R2−−−−−−−−−→
1 −2 10 1 50 3 −5
∣∣∣∣∣∣3−1−3
Jason Aubrey Math 1300 Finite Mathematics
![Page 34: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/34.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
−3R1+R2→R2−−−−−−−−−→
1 −2 10 7 −52 −1 −3
∣∣∣∣∣∣3−73
−2R1+R3→R3−−−−−−−−−→
1 −2 10 7 −50 3 −5
∣∣∣∣∣∣3−7−3
−2R3+R2→R2−−−−−−−−−→
1 −2 10 1 50 3 −5
∣∣∣∣∣∣3−1−3
Jason Aubrey Math 1300 Finite Mathematics
![Page 35: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/35.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
−3R1+R2→R2−−−−−−−−−→
1 −2 10 7 −52 −1 −3
∣∣∣∣∣∣3−73
−2R1+R3→R3−−−−−−−−−→
1 −2 10 7 −50 3 −5
∣∣∣∣∣∣3−7−3
−2R3+R2→R2−−−−−−−−−→
1 −2 10 1 50 3 −5
∣∣∣∣∣∣3−1−3
Jason Aubrey Math 1300 Finite Mathematics
![Page 36: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/36.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
2R2+R1→R1−−−−−−−−→
1 0 110 1 50 3 −5
∣∣∣∣∣∣1−1−3
−3R2+R3→R3−−−−−−−−−→
1 0 110 1 50 0 −20
∣∣∣∣∣∣1−10
− 1
20 R3→R3−−−−−−−→
1 0 110 1 50 0 1
∣∣∣∣∣∣1−10
−5R3+R2→R2−−−−−−−−−→
Jason Aubrey Math 1300 Finite Mathematics
![Page 37: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/37.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
2R2+R1→R1−−−−−−−−→
1 0 110 1 50 3 −5
∣∣∣∣∣∣1−1−3
−3R2+R3→R3−−−−−−−−−→
1 0 110 1 50 0 −20
∣∣∣∣∣∣1−10
− 1
20 R3→R3−−−−−−−→
1 0 110 1 50 0 1
∣∣∣∣∣∣1−10
−5R3+R2→R2−−−−−−−−−→
Jason Aubrey Math 1300 Finite Mathematics
![Page 38: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/38.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
2R2+R1→R1−−−−−−−−→
1 0 110 1 50 3 −5
∣∣∣∣∣∣1−1−3
−3R2+R3→R3−−−−−−−−−→
1 0 110 1 50 0 −20
∣∣∣∣∣∣1−10
− 1
20 R3→R3−−−−−−−→
1 0 110 1 50 0 1
∣∣∣∣∣∣1−10
−5R3+R2→R2−−−−−−−−−→
Jason Aubrey Math 1300 Finite Mathematics
![Page 39: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/39.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
2R2+R1→R1−−−−−−−−→
1 0 110 1 50 3 −5
∣∣∣∣∣∣1−1−3
−3R2+R3→R3−−−−−−−−−→
1 0 110 1 50 0 −20
∣∣∣∣∣∣1−10
− 120 R3→R3−−−−−−−→
1 0 110 1 50 0 1
∣∣∣∣∣∣1−10
−5R3+R2→R2−−−−−−−−−→
Jason Aubrey Math 1300 Finite Mathematics
![Page 40: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/40.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
2R2+R1→R1−−−−−−−−→
1 0 110 1 50 3 −5
∣∣∣∣∣∣1−1−3
−3R2+R3→R3−−−−−−−−−→
1 0 110 1 50 0 −20
∣∣∣∣∣∣1−10
− 1
20 R3→R3−−−−−−−→
1 0 110 1 50 0 1
∣∣∣∣∣∣1−10
−5R3+R2→R2−−−−−−−−−→
Jason Aubrey Math 1300 Finite Mathematics
![Page 41: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/41.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
2R2+R1→R1−−−−−−−−→
1 0 110 1 50 3 −5
∣∣∣∣∣∣1−1−3
−3R2+R3→R3−−−−−−−−−→
1 0 110 1 50 0 −20
∣∣∣∣∣∣1−10
− 1
20 R3→R3−−−−−−−→
1 0 110 1 50 0 1
∣∣∣∣∣∣1−10
−5R3+R2→R2−−−−−−−−−→
Jason Aubrey Math 1300 Finite Mathematics
![Page 42: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/42.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
2R2+R1→R1−−−−−−−−→
1 0 110 1 50 3 −5
∣∣∣∣∣∣1−1−3
−3R2+R3→R3−−−−−−−−−→
1 0 110 1 50 0 −20
∣∣∣∣∣∣1−10
− 1
20 R3→R3−−−−−−−→
1 0 110 1 50 0 1
∣∣∣∣∣∣1−10
−5R3+R2→R2−−−−−−−−−→
Jason Aubrey Math 1300 Finite Mathematics
![Page 43: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/43.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
1 0 110 1 00 0 1
∣∣∣∣∣∣1−10
−11R3+R1→R1−−−−−−−−−−→
1 0 00 1 00 0 1
∣∣∣∣∣∣1−10
The solution is therefore x1 = 1, x2 = −1 and x3 = 0.
Jason Aubrey Math 1300 Finite Mathematics
![Page 44: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/44.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
1 0 110 1 00 0 1
∣∣∣∣∣∣1−10
−11R3+R1→R1−−−−−−−−−−→
1 0 00 1 00 0 1
∣∣∣∣∣∣1−10
The solution is therefore x1 = 1, x2 = −1 and x3 = 0.
Jason Aubrey Math 1300 Finite Mathematics
![Page 45: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/45.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
1 0 110 1 00 0 1
∣∣∣∣∣∣1−10
−11R3+R1→R1−−−−−−−−−−→
1 0 00 1 00 0 1
∣∣∣∣∣∣1−10
The solution is therefore x1 = 1, x2 = −1 and x3 = 0.
Jason Aubrey Math 1300 Finite Mathematics
![Page 46: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/46.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
1 0 110 1 00 0 1
∣∣∣∣∣∣1−10
−11R3+R1→R1−−−−−−−−−−→
1 0 00 1 00 0 1
∣∣∣∣∣∣1−10
The solution is therefore x1 = 1, x2 = −1 and x3 = 0.
Jason Aubrey Math 1300 Finite Mathematics
![Page 47: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/47.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Definition (Gauss-Jordan Elimination)
1 Choose the leftmost nonzero column and use appropriaterow operations to get a 1 at the top.
2 Use multiples of the row containing the 1 from step 1 to getzeros in all remaining places in the column containing the1.
3 Repeat step 1 with the submatrix formed by (mentally)deleting the row used in step 2 and all rows above that row.
4 Repeat step 2 with the entire matrix, including the rowsdeleted mentally. Continue this process until the entirematrix is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 48: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/48.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Definition (Gauss-Jordan Elimination)1 Choose the leftmost nonzero column and use appropriate
row operations to get a 1 at the top.
2 Use multiples of the row containing the 1 from step 1 to getzeros in all remaining places in the column containing the1.
3 Repeat step 1 with the submatrix formed by (mentally)deleting the row used in step 2 and all rows above that row.
4 Repeat step 2 with the entire matrix, including the rowsdeleted mentally. Continue this process until the entirematrix is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 49: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/49.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Definition (Gauss-Jordan Elimination)1 Choose the leftmost nonzero column and use appropriate
row operations to get a 1 at the top.2 Use multiples of the row containing the 1 from step 1 to get
zeros in all remaining places in the column containing the1.
3 Repeat step 1 with the submatrix formed by (mentally)deleting the row used in step 2 and all rows above that row.
4 Repeat step 2 with the entire matrix, including the rowsdeleted mentally. Continue this process until the entirematrix is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 50: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/50.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Definition (Gauss-Jordan Elimination)1 Choose the leftmost nonzero column and use appropriate
row operations to get a 1 at the top.2 Use multiples of the row containing the 1 from step 1 to get
zeros in all remaining places in the column containing the1.
3 Repeat step 1 with the submatrix formed by (mentally)deleting the row used in step 2 and all rows above that row.
4 Repeat step 2 with the entire matrix, including the rowsdeleted mentally. Continue this process until the entirematrix is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 51: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/51.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Definition (Gauss-Jordan Elimination)1 Choose the leftmost nonzero column and use appropriate
row operations to get a 1 at the top.2 Use multiples of the row containing the 1 from step 1 to get
zeros in all remaining places in the column containing the1.
3 Repeat step 1 with the submatrix formed by (mentally)deleting the row used in step 2 and all rows above that row.
4 Repeat step 2 with the entire matrix, including the rowsdeleted mentally. Continue this process until the entirematrix is in reduced form.
Jason Aubrey Math 1300 Finite Mathematics
![Page 52: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/52.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]3R1+R2→R2−−−−−−−−→[
1 2 −10 12 −11
∣∣∣∣ 1−1
]1
12 R2→R2−−−−−−→[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]−2R2+R1→R1−−−−−−−−−→[
1 0 1012
0 1 −1112
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 53: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/53.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]
12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]3R1+R2→R2−−−−−−−−→[
1 2 −10 12 −11
∣∣∣∣ 1−1
]1
12 R2→R2−−−−−−→[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]−2R2+R1→R1−−−−−−−−−→[
1 0 1012
0 1 −1112
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 54: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/54.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]3R1+R2→R2−−−−−−−−→[
1 2 −10 12 −11
∣∣∣∣ 1−1
]1
12 R2→R2−−−−−−→[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]−2R2+R1→R1−−−−−−−−−→[
1 0 1012
0 1 −1112
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 55: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/55.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]
3R1+R2→R2−−−−−−−−→[1 2 −10 12 −11
∣∣∣∣ 1−1
]1
12 R2→R2−−−−−−→[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]−2R2+R1→R1−−−−−−−−−→[
1 0 1012
0 1 −1112
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 56: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/56.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]3R1+R2→R2−−−−−−−−→
[1 2 −10 12 −11
∣∣∣∣ 1−1
]1
12 R2→R2−−−−−−→[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]−2R2+R1→R1−−−−−−−−−→[
1 0 1012
0 1 −1112
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 57: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/57.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]3R1+R2→R2−−−−−−−−→[
1 2 −10 12 −11
∣∣∣∣ 1−1
]
112 R2→R2−−−−−−→
[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]−2R2+R1→R1−−−−−−−−−→[
1 0 1012
0 1 −1112
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 58: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/58.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]3R1+R2→R2−−−−−−−−→[
1 2 −10 12 −11
∣∣∣∣ 1−1
]1
12 R2→R2−−−−−−→
[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]−2R2+R1→R1−−−−−−−−−→[
1 0 1012
0 1 −1112
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 59: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/59.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]3R1+R2→R2−−−−−−−−→[
1 2 −10 12 −11
∣∣∣∣ 1−1
]1
12 R2→R2−−−−−−→[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]
−2R2+R1→R1−−−−−−−−−→[1 0 10
120 1 −11
12
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 60: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/60.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]3R1+R2→R2−−−−−−−−→[
1 2 −10 12 −11
∣∣∣∣ 1−1
]1
12 R2→R2−−−−−−→[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]−2R2+R1→R1−−−−−−−−−→
[1 0 10
120 1 −11
12
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 61: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/61.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
2x1 + 4x2 − 2x3 = 2−3x1 + 6x2 − 8x3 = −4
[2 4 −2−3 6 −8
∣∣∣∣ 2−4
]12 R1→R1−−−−−→
[1 2 −1−3 6 −8
∣∣∣∣ 1−4
]3R1+R2→R2−−−−−−−−→[
1 2 −10 12 −11
∣∣∣∣ 1−1
]1
12 R2→R2−−−−−−→[1 2 −10 1 −11
12
∣∣∣∣ 1− 1
12
]−2R2+R1→R1−−−−−−−−−→[
1 0 1012
0 1 −1112
∣∣∣∣ 1412− 1
12
]
Jason Aubrey Math 1300 Finite Mathematics
![Page 62: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/62.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
This last matrix corresponds to the system of equations
x1 +56x3 = 7
6x2 −11
12x3 = − 112
So we set
x1 =76− 5
6t
x2 = − 112
+1112
t
x3 = t
Jason Aubrey Math 1300 Finite Mathematics
![Page 63: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/63.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
This last matrix corresponds to the system of equations
x1 +56x3 = 7
6x2 −11
12x3 = − 112
So we set
x1 =76− 5
6t
x2 = − 112
+1112
t
x3 = t
Jason Aubrey Math 1300 Finite Mathematics
![Page 64: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/64.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
This last matrix corresponds to the system of equations
x1 +56x3 = 7
6x2 −11
12x3 = − 112
So we set
x1 =76− 5
6t
x2 = − 112
+1112
t
x3 = t
Jason Aubrey Math 1300 Finite Mathematics
![Page 65: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/65.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
This last matrix corresponds to the system of equations
x1 +56x3 = 7
6x2 −11
12x3 = − 112
So we set
x1 =76− 5
6t
x2 = − 112
+1112
t
x3 = t
Jason Aubrey Math 1300 Finite Mathematics
![Page 66: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/66.jpg)
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Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
4x1 − 2x2 + 2x3 = 5−6x1 + 3x2 − 3x3 = −2
10x1 − 5x2 − x3 = −19
This system has no solutions because,
4 −2 2−6 3 −310 −5 −1
∣∣∣∣∣∣5−2−19
14 R1→R1−−−−−→
1 −12
12
−6 3 −310 −5 −1
∣∣∣∣∣∣54−2−19
6R1+R2→R2−−−−−−−−→ 1 −12
12
0 0 010 −5 −1
∣∣∣∣∣∣54
112−19
Jason Aubrey Math 1300 Finite Mathematics
![Page 67: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/67.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
4x1 − 2x2 + 2x3 = 5−6x1 + 3x2 − 3x3 = −2
10x1 − 5x2 − x3 = −19
This system has no solutions because,
4 −2 2−6 3 −310 −5 −1
∣∣∣∣∣∣5−2−19
14 R1→R1−−−−−→
1 −12
12
−6 3 −310 −5 −1
∣∣∣∣∣∣54−2−19
6R1+R2→R2−−−−−−−−→ 1 −12
12
0 0 010 −5 −1
∣∣∣∣∣∣54
112−19
Jason Aubrey Math 1300 Finite Mathematics
![Page 68: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/68.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
4x1 − 2x2 + 2x3 = 5−6x1 + 3x2 − 3x3 = −2
10x1 − 5x2 − x3 = −19
This system has no solutions because,
4 −2 2−6 3 −310 −5 −1
∣∣∣∣∣∣5−2−19
14 R1→R1−−−−−→
1 −12
12
−6 3 −310 −5 −1
∣∣∣∣∣∣54−2−19
6R1+R2→R2−−−−−−−−→ 1 −12
12
0 0 010 −5 −1
∣∣∣∣∣∣54
112−19
Jason Aubrey Math 1300 Finite Mathematics
![Page 69: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/69.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
4x1 − 2x2 + 2x3 = 5−6x1 + 3x2 − 3x3 = −2
10x1 − 5x2 − x3 = −19
This system has no solutions because,
4 −2 2−6 3 −310 −5 −1
∣∣∣∣∣∣5−2−19
14 R1→R1−−−−−→
1 −12
12
−6 3 −310 −5 −1
∣∣∣∣∣∣54−2−19
6R1+R2→R2−−−−−−−−→ 1 −12
12
0 0 010 −5 −1
∣∣∣∣∣∣54
112−19
Jason Aubrey Math 1300 Finite Mathematics
![Page 70: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/70.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
4x1 − 2x2 + 2x3 = 5−6x1 + 3x2 − 3x3 = −2
10x1 − 5x2 − x3 = −19
This system has no solutions because,
4 −2 2−6 3 −310 −5 −1
∣∣∣∣∣∣5−2−19
14 R1→R1−−−−−→
1 −12
12
−6 3 −310 −5 −1
∣∣∣∣∣∣54−2−19
6R1+R2→R2−−−−−−−−→
1 −12
12
0 0 010 −5 −1
∣∣∣∣∣∣54
112−19
Jason Aubrey Math 1300 Finite Mathematics
![Page 71: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/71.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: Solve the system using Gauss-Jordan elimination:
4x1 − 2x2 + 2x3 = 5−6x1 + 3x2 − 3x3 = −2
10x1 − 5x2 − x3 = −19
This system has no solutions because,
4 −2 2−6 3 −310 −5 −1
∣∣∣∣∣∣5−2−19
14 R1→R1−−−−−→
1 −12
12
−6 3 −310 −5 −1
∣∣∣∣∣∣54−2−19
6R1+R2→R2−−−−−−−−→ 1 −12
12
0 0 010 −5 −1
∣∣∣∣∣∣54
112−19
Jason Aubrey Math 1300 Finite Mathematics
![Page 72: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/72.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: A fruit grower can use two types of fertilizer in anorange grove, brand A and brand B. Each bag of brand Acontains 8 pounds of nitrogen and 4 pounds of phosphoric acid.Each bag of brand B contains 7 pounds of nitrogen and 6pounds of phosphoric acid. Tests indicate that the grove needs720 pounds of nitrogen and 500 pounds of phosphoric acid.How many bags of each brand should be used to provide therequired amounts of nitrogen and phosphoric acid?
Jason Aubrey Math 1300 Finite Mathematics
![Page 73: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/73.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We begin by organizing the data into a table:
Brand A Brand BNitrogen 8 lbs 7 lbs
Phosphoric Acid 4 lbs 6 lbs
Next, we assign variables for each of the unknowns.
x1 = # of bags of Brand Ax2 = # of bags of Brand B
Jason Aubrey Math 1300 Finite Mathematics
![Page 74: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/74.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We begin by organizing the data into a table:
Brand A Brand BNitrogen 8 lbs 7 lbs
Phosphoric Acid 4 lbs 6 lbs
Next, we assign variables for each of the unknowns.
x1 = # of bags of Brand Ax2 = # of bags of Brand B
Jason Aubrey Math 1300 Finite Mathematics
![Page 75: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/75.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We begin by organizing the data into a table:
Brand A Brand BNitrogen 8 lbs 7 lbs
Phosphoric Acid 4 lbs 6 lbs
Next, we assign variables for each of the unknowns.
x1 = # of bags of Brand Ax2 = # of bags of Brand B
Jason Aubrey Math 1300 Finite Mathematics
![Page 76: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/76.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 77: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/77.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 78: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/78.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
]
18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 79: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/79.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 80: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/80.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]
−4R1+R2→R2−−−−−−−−−→[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 81: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/81.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 82: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/82.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]
25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 83: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/83.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 84: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/84.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]
− 78 R2+R1→R1−−−−−−−−−→
[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 85: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/85.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→
[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 86: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/86.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 87: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/87.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
We now set up our linear equations and solve.
8x1 + 7x2 = 7204x1 + 6x2 = 500
[8 74 6
∣∣∣∣720500
] 18 R1→R1−−−−−→
[1 7
84 6
∣∣∣∣ 90500
]−4R1+R2→R2−−−−−−−−−→
[1 7
80 5
2
∣∣∣∣ 90140
]25 R2→R2−−−−−→
[1 7
80 1
∣∣∣∣9056
]− 7
8 R2+R1→R1−−−−−−−−−→[1 00 1
∣∣∣∣4156
]
So, x1 = 41 and x2 = 56.
Jason Aubrey Math 1300 Finite Mathematics
![Page 88: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/88.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: A fruit grower can use two types of fertilizer in anorange grove, brand A and brand B. Each bag of brand Acontains 8 pounds of nitrogen and 4 pounds of phosphoric acid.Each bag of brand B contains 7 pounds of nitrogen and 6pounds of phosphoric acid. Tests indicate that the grove needs720 pounds of nitrogen and 500 pounds of phosphoric acid.How many bags of each brand should be used to provide therequired amounts of nitrogen and phosphoric acid?
Answer: The grower should use 41 bags of brand A and 56bags of brand B.
Jason Aubrey Math 1300 Finite Mathematics
![Page 89: Math 1300: Section 4- 3 Gauss-Jordan Elimination](https://reader031.vdocument.in/reader031/viewer/2022020206/5482f0eab47959ec0c8b4951/html5/thumbnails/89.jpg)
university-logo
Reduced MatricesSolving Systems by Gauss-Jordan Elimination
Application
Example: A fruit grower can use two types of fertilizer in anorange grove, brand A and brand B. Each bag of brand Acontains 8 pounds of nitrogen and 4 pounds of phosphoric acid.Each bag of brand B contains 7 pounds of nitrogen and 6pounds of phosphoric acid. Tests indicate that the grove needs720 pounds of nitrogen and 500 pounds of phosphoric acid.How many bags of each brand should be used to provide therequired amounts of nitrogen and phosphoric acid?
Answer: The grower should use 41 bags of brand A and 56bags of brand B.
Jason Aubrey Math 1300 Finite Mathematics