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Row Reducing How  to get to Reduced Row Echelon Form
Strategies ,[object Object]
  Get the leading 1’s starting at the pivot
  Use the pivot to get zeros below  and/or above ,[object Object],      other as you move right
Elementary Row Operations Allowable Row Operations Notation Ri↔Rj ,[object Object]
  Multiply a row through       by a non-zero constant ,[object Object],      row to another.  kRi->Ri kRi + Rj-> Rj
EXAMPLE I PIVOT Use this one, called the pivot, to get zeros below. -3R1 + R2-> R2 R1 + R3-> R3
EXAMPLE I -3R1 + R2-> R2
EXAMPLE I R1 + R3-> R3
EXAMPLE I New pivot location We want a 1 here but we also want to avoid fractions -2R3 + R2-> R2
EXAMPLE I New pivot  Use this new pivot to get zeros above and below    R2 + R1-> R1 -2R2 + R3-> R3
EXAMPLE I New pivot location  Now get a 1 here.       R3-> R3
EXAMPLE I New pivot Use this new pivot to make zeros above. 11R3 + R1-> R1 13R3 + R2-> R2
EXAMPLE I
EXAMPLE  II Solve the system:  There are two equations and four unknowns so we need a 2 x 5 matrix. Augmented matrix:
EXAMPLE  II -2R1 + R2-> R2     R2-> R2 ←RREF

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Row Reducing

  • 1. Row Reducing How to get to Reduced Row Echelon Form
  • 2.
  • 3. Get the leading 1’s starting at the pivot
  • 4.
  • 5.
  • 6.
  • 7. EXAMPLE I PIVOT Use this one, called the pivot, to get zeros below. -3R1 + R2-> R2 R1 + R3-> R3
  • 8. EXAMPLE I -3R1 + R2-> R2
  • 9. EXAMPLE I R1 + R3-> R3
  • 10. EXAMPLE I New pivot location We want a 1 here but we also want to avoid fractions -2R3 + R2-> R2
  • 11. EXAMPLE I New pivot Use this new pivot to get zeros above and below R2 + R1-> R1 -2R2 + R3-> R3
  • 12. EXAMPLE I New pivot location Now get a 1 here. R3-> R3
  • 13. EXAMPLE I New pivot Use this new pivot to make zeros above. 11R3 + R1-> R1 13R3 + R2-> R2
  • 15. EXAMPLE II Solve the system: There are two equations and four unknowns so we need a 2 x 5 matrix. Augmented matrix:
  • 16. EXAMPLE II -2R1 + R2-> R2 R2-> R2 ←RREF
  • 17. EXAMPLE II Expressing the solution: If a system is consistent, any column that does not contain a leading one will have a parameter. Parameters required for x3 and x4
  • 18. EXAMPLE II Expressing the solution: LET s t
  • 19. EXAMPLE II Converting back to equations in terms of s and t: Solving for x1 and x2 leads to the final solution:

Notes de l'éditeur

  1. 1.2C