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university-logo Reduced Matrices Solving Systems by Gauss-Jordan Elimination Application Math 1300 Finite Mathematics Section 4.3 Gauss-Jordan Elimination Jason Aubrey Department of Mathematics University of Missouri Jason Aubrey Math 1300 Finite Mathematics

Math 1300: Section 4- 3 Gauss-Jordan Elimination

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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

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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

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

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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