B œ c " " ã B œ c 8 8. such that substituting these values for the B 3 's will make all the equations true

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Rose Booth
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1 System of Linear Equations variables Ð unknowns Ñ B" ß B# ß ÞÞÞ ß B8 Æ Æ Æ + B + B ÞÞÞ + B œ, "" " "# # "8 8 " + B + B ÞÞÞ + B œ, #" " ## # #8 8 # ã + B + B ÞÞÞ + B œ, 3" " 3# # ã + 7" B" + 7# B# ÞÞÞ + 78B8 œ, 7 Å Å Å constant coefficients ( real or complex numbers) A solution means a list of numbers B œ c " " ã B œ c 8 8 such that substituting these values for the B 3 's will make all the equations true We can summarize the system of equations using the augmented matrix Ã augmented matrix Ä Ô + "" + "# ÞÞÞ + "8 l," + #" + ## ÞÞÞ + #8 l, # l a3" + 3# ÞÞÞ + 38 l, 3 Ö Ù ÞÞÞ l Õ+ + + Ø 7" 7# 78 l, 7 â vertical line unofficial, to emphasize, 3 's are not coefficients in the equations Ã coefficient matrix Ä ÐThe matrix with containing just the blue entries is called the coefficient matrix of the system) Here, our focus is on the rows of the matrix because each row represents an equation

2 Most important questions about any linear system of equations 1) Is there a solution (does a solution exist)? yes Ç consistent system no Ç inconsistent system Ô B" Ô -" B# -# A solution could be written in form Ö Ù œ Ö Ù ã ã ÕB Ø Õ- Ø 2) If there is a solution (consistent system) 7 7 a) is there unique solution (only one solution)? b) is there more than one solution? For a linear system of equations, it turns out that if there is more than one solution, then there must be an infinitely many solutions. For example, it's impossible for a linear system to have exactly 2 (different) solutions or Ô B" Ô -" Ô B " Ô." B# -# B #.# such asö Ù œ Ö Ù and Ö Ù œ Ö Ù ã ã ã ã ÕB Ø Õ- Ø ÕB Ø Õ. Ø

3 Certain operations on system of equation Ò new system with same solutions: an equivalent Ð µ Ñ system For example, the operations below lead to an equivalent system: (interchange eqns) œ %B " #B # œ ' œ #B " #B # µ œ # #B #B œ # %B #B œ ' " # " # add multiple of one eqn multiply eqns by nonzero to another Æ constant Æ #B" #B# œ # B" B# œ " µ œ µ! #B œ "! œ! B œ & # # add multiple of one equation to another Æ B! œ % µ œ "! B# œ & We can abbreviate by writing only the augmented matrix, omitting the variables. We perform the operations on the rows of the matrix. % # ' µ # # # # # # % # ' # # # " " " µ µ! # "!! " & "! % µ! " & The new (equivalent) system exhibits the solutions very clearly: B œ % " B œ & #

4 For each equation above, the set of all ÐBßCÑ that satisfy the equation (the solution set ) is a straight line. The two lines here intersect in a single point, Ð %ß &Ñ, that is the unique solution for the system. Graph illustrates: the lines represented by the two equations have exactly one intersection, at ÐB ßB Ñ œ Ð %ß &Ñ " # Arguing graphically in a similar way: any system of two linear equations with 2 unknowns has either no solution, a unique solution, or infinitely many solutions. (Describe the geometry in each case) Geometrically, explain why the system #B $C %D œ & &B (C $B œ "# must have either no solution or infinitely many solutions.

5 What standard operations on a system that produce an equivalent system (one with same solutions)? We call them Elementary Row Operations ( EROs) 1) interchange any two rows ( œ swap rows ) 2) multiply any row by a nonzero constant - ( œ rescale rowñ 3) add a multiple of one row to another row ( œ row replacement: a row is replaced by the result of adding another row to it) Two matrices (or, systems of equations) are called row equivalent if you can change one into the other by using a sequence of EROs Important: Each ERO operation is reversible (can be undone ) by using some other ERO Swap: interchange row 1 (R1) and row 2 (R2) What ERO reverses the first? Æ Æ Ô " # $ Ô % & ' Ô % & ' { " # $ { Õ( ) * Ø Õ( ) * Ø Õ 7 8 9Ø Rescale: multiply R3 by & What ERO reverses the first? Æ Æ Ô " # $ Ô " # $ Ô " # $ % & ' { % & ' { % & ' Õ( ) * Ø Õ$& %! %& Ø Õ( ) * Ø Row Replacement: add 3*R1 to R3 What ERO reverses the first? Æ Æ Ô " # $ Ô " # $ Ô " # $ % & ' { % & ' { % & ' Õ( ) * Ø Õ"! "% ") Ø Õ( ) * Ø

6 The point is that EROs, correctly chosen, will change the system into an equivalent system where it's easy to read off the solutions B B B œ! " # $ #B $B 'B œ "# " # $ Aug. Matrix Æ " " "! µ " " "! µ " " "! # $ ' "#! & % "#! " %Î& "#Î& µ "! *Î& "#Î&! " %Î& "#Î& Å Å Å Å B" B# B$ constants so can write "# * B" œ & & B$ andb$ is free (can have "# % B# œ B & & $ any value)

5x 2 = 10. x 1 + 7(2) = 4. x 1 3x 2 = 4. 3x 1 + 9x 2 = 8

5x 2 = 10. x 1 + 7(2) = 4. x 1 3x 2 = 4. 3x 1 + 9x 2 = 8 1 To solve the system x 1 + x 2 = 4 2x 1 9x 2 = 2 we find an (easier to solve) equivalent system as follows: Replace equation 2 with (2 times equation 1 + equation 2): x 1 + x 2 = 4 Solve equation 2 for