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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
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
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
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
university-logo
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
university-logo
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
university-logo
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
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
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
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
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
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
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
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
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
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
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
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
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
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
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
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
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
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
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
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
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
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
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
university-logo
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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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
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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
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