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Multivariable Linear Systems
Skill 21
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Objectives Use Back-Substitution to solve systems in Row-Echelon Form
Use Gaussian Elimination to solve systems of linear equations Solve non-square systems of linear equations Find partial fraction decomposition
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Row-Echelon A form to put systems in so you can quickly use substitution to solve. Modify the system until the first equation has three terms. Modify the system until the second equation has two terms. Modify the system until the third equation has one term. Substitute backwards one at a time.
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Example; Solve Row-Echelon Form
2๐ฅโ๐ฆ+3๐ง=17 โ7๐ฆโ4๐ง=โ9 ๐ง=4 โ7๐ฆโ4 ๐ง =โ9 2๐ฅโ๐ฆ+3๐ง=17 โ7๐ฆโ4 4 =โ9 2๐ฅโ โ =17 โ7๐ฆ=7 2๐ฅ=4 ๐=โ๐ ๐=๐ ๐,โ๐,๐
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Example; Solve Row-Echelon Form
โ3๐ฅ+2๐ฆโ4๐ง=15 5๐ฆโ2๐ง=27 ๐ง=โ6 5๐ฆโ2 ๐ง =27 โ3๐ฅ+2๐ฆโ4๐ง=15 5๐ฆโ2 โ6 =27 โ3๐ฅโ 3 +3 โ6 =15 5๐ฆ=15 โ3๐ฅ=โ15 ๐=๐ ๐=๐ ๐,๐,โ๐
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Gaussian Elimination Elementary Row Operations for Systems of Equations 1) Interchange two equations. 2) Multiply one equation by a non-zero constant. 3) Add a multiple of one equation to another equation.
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Example; Using Gaussian Elimination
๐ฅโ2๐ฆ+3๐ง=9 ๐ฆ+4๐ง=7 2๐ฅโ5๐ฆ+5๐ง=17 ๐น ๐ + ๐น ๐ ๐ฅโ2๐ฆ+3๐ง=9 ๐ฆ+4๐ง=7 โ๐ฆโ๐ง=โ1 ๐โ๐๐+๐๐=๐ โ๐+๐๐+๐=โ๐ ๐๐โ๐๐+๐๐=๐๐ ๐ ๐น ๐ + ๐น ๐ ๐ฅโ2๐ฆ+3๐ง=9 ๐ฆ+4๐ง=7 3๐ง=6 ๐น ๐ + ๐น ๐ ๐โ๐๐+๐๐=๐ ๐+๐๐=๐ ๐=๐ ๐ ๐ ๐น ๐
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Example; Using Gaussian Elimination, Continued
๐ฅโ2๐ฆ+3๐ง=9 ๐ฆ+4๐ง=7 ๐ง=2 ๐ฆ+4 ๐ง =7 ๐ฅโ2๐ฆ+3๐ง=9 ๐ฆ+4 2 =7 ๐ฅโ2 โ =9 ๐ฆ+8=7 ๐ฅ+2+6=9 ๐=๐ ๐=โ๐ ๐,โ๐,๐
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Example; Using Gaussian Elimination
๐ฅ+๐ฆ+๐ง=6 โ3๐ฆโ๐ง=โ9 โ2๐ฆโ4๐ง=โ16 โ๐ ๐น ๐ + ๐น ๐ โ๐ ๐น ๐ + ๐น ๐ ๐ฅ+๐ฆ+๐ง=6 ๐ฆ+2๐ง=8 โ3๐ฆโ๐ง=โ9 โ ๐ ๐ ๐น ๐ โ ๐น ๐ ๐+๐+๐=๐ ๐๐โ๐+๐=๐ ๐๐+๐โ๐=๐ ๐ฅ+๐ฆ+๐ง=6 ๐ฆ+2๐ง=8 5๐ง=15 ๐ ๐น ๐ + ๐น ๐ ๐+๐+๐=๐ ๐+๐๐=๐ ๐=๐ ๐ ๐ ๐น ๐
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Example; Using Gaussian Elimination, Continued
๐ฅ+๐ฆ+๐ง=6 ๐ฆ+2๐ง=8 ๐ง=3 ๐ฆ+2 ๐ง =8 ๐ฅ+๐ฆ+๐ง=6 ๐ฆ+2 3 =8 ๐ฅ =6 ๐ฆ+6=8 ๐ฅ+5=6 ๐=๐ ๐=๐ ๐,๐,๐
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Example; Systems with fewer equations than variables
๐ฅโ2๐ฆ+๐ง=2 3๐ฆโ3๐ง=โ3 โ๐ ๐น ๐ + ๐น ๐ ๐โ๐๐+๐=๐ ๐๐โ๐โ๐=๐ ๐ฅโ2๐ฆ+๐ง=2 ๐ฆโ๐ง=โ1 ๐ ๐ ๐น ๐ ๐ฅโ2๐ฆ+๐ง=2 ๐ฆ=๐งโ1 ๐ฅโ2๐ฆ+๐ง=2 ๐ฅโ2 ๐งโ1 +๐ง=2 Let ๐=๐ ๐ฅโ2๐ง+2+๐ง=2 So, ๐=๐ ๐=๐ and, ๐=๐โ๐ ๐,๐โ๐,๐
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Example; Systems with fewer equations than variables
๐โ๐+๐๐=๐ ๐๐โ๐=๐ Let ๐=๐ ๐ฅโ๐ฆ+4๐ง=2 ๐ง=4๐ฅ So, ๐=๐๐ and, ๐=๐๐โ๐ ๐ฅโ๐ฆ+4๐ง=3 ๐ฅโ๐ฆ+4๐ฅ=3 5๐ฅโ๐ฆ=3 ๐,๐๐โ๐,๐๐ ๐=๐๐โ๐
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Partial Fraction Decomposition
1) Divide if improper 2) Factor Denominator 3) Linear Factors 4) Quadratic Factors
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Example; Partial Fraction Decomposition
๐ฅ+7 ๐ฅ 2 โ๐ฅโ6 = ๐ด ๐ฅโ3 + ๐ต ๐ฅ+2 ๐ฅ+7=๐ด ๐ฅ+2 +๐ต ๐ฅโ3 ๐ฅ+7=๐ด๐ฅ+2๐ด+๐ต๐ฅโ3๐ต 1=๐ด+๐ต 7=2๐ดโ3๐ต โ๐ 1=๐ด+๐ต โ2=โ2๐ดโ2๐ต 7=2๐ดโ3๐ต 1=๐ดโ1 + ๐จ=๐ ๐ ๐โ๐ โ ๐ ๐+๐ 5=โ5๐ต ๐ฉ=โ๐
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Example; Partial Fraction Decomposition
๐ฅ+8 ๐ฅ 2 +6๐ฅ+8 = ๐ด ๐ฅ+4 + ๐ต ๐ฅ+2 ๐ฅ+8=๐ด ๐ฅ+2 +๐ต ๐ฅ+4 ๐ฅ+8=๐ด๐ฅ+2๐ด+๐ต๐ฅ+4๐ต 1=๐ด+๐ต 8=2๐ด+4๐ต โ๐ 1=๐ด+๐ต โ2=โ2๐ดโ2๐ต 8=2๐ด+4๐ต 1=๐ด+3 + ๐จ=โ๐ 6=2๐ต โ๐ ๐+๐ + ๐ ๐+๐ ๐ฉ=๐
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21: Multivariable Linear Systems
Summarize your notes Questions? Homework Worksheet Quiz
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