Elementary Row Operations (s)ERO for a matrix (or a system of linear equations) 1) interchange any two equations (rows) 2) multiply any equation (row) by a nonzero constant - 3) add a multiple of one equation (row) to another equation (row) Each operation is reversible that is, can be “undone” by using another ERO. _____________________________________________________ Example BB"# Bœ" $ #B"#$ $B B œ % In matrix notation: B#BBœ$"#$ """ " Augmented matrix œ # $ " % " # " $ """ " µ What ERO? ! " $ ' " # " $ "" " " µ What ERO? ! " $ ' !$ # % "" " " µ What ERO? ! " $ ' ! ! "" ## "" " " µ What ERO? ! " $ ' !! " # """" µ What ERO? !"!! !!"# ""! " µ What ERO? !"! ! !!" # "!! " µ What ERO? !"! ! !!" # which means: Bœ" ? Bœ ? # Bœ ? $ __________________________________________________ Example B"#$ $B %B œ % $B"#$ (B (B œ ) In matrix notation %B"#$ 'B B œ ( "$%% $(() % ' " ( "$%% µ What ERO? !#&% % ' " ( "$%% µ What ERO? !#&% !'"&* "$ % % µ What ERO?! " &Î# # !'"&* "$% % µ What ERO? ! " &Î# # so ... ? !! !$ EXAMPLE (See Section 1.6) Chemical Reaction boron sulfide + water --> boric acid + hydrogen sulfide ÐB"# ÑBS#$+ ÐB Ñ HO # ----> ÐB $ Ñ HBO $ $+ ÐB % Ñ HS # Same number of boron atoms on both sides so: #B"$ œ B Same number of sulfur atoms on both sides: $B"% œ B Same number of hydrogen atoms on both sides: #B#$% œ $B #B Same number of oxygen atoms on both sides: Bœ$B#$ To satisfy all these equations: #B"$ B œ ! $B"% B œ ! #B#$% $B #B œ ! B$B#$ œ! Row Reduction Steps (Identify the ERO at each step) #! " !! "! "Î# !! $! ! "! $! ! "! µ ! # $ # ! ! # $ # ! !" $ !! !" $ !! "! "Î# !! "! "Î# !! ! ! $Î# " ! ! " $ ! ! µµ !# $ #! !# $#! !" $ !! !! $Î#"! "! "Î# !! "! "Î# !! !" $ !! !" $ !! µµ !! $ #! !! $ #! ! ! $Î# " ! ! ! ! ! ! " ! "Î# ! ! " ! "Î# ! ! !" $ ! ! !" ! #! µµ ! ! " #Î$ ! ! ! " #Î$ ! !! ! ! ! !! ! ! ! " ! ! "Î$ ! !"! #! µ or ! ! " #Î$ ! !!! ! ! " BBœ!"% $ B#Bœ!#% # so BBœ!$%$ !œ! " Bœ"%$ B, Bœ#B#%, # Bœ$%$ B where B % can have any value Ðx% is a free variable) mathematically. So: infinitely many solutions). For example, let Bœ$% (chemically , we need postive integ / r solutions). then Bœ"ßBœ'ßBœ#"#$, and B#$ S + '#$ H # O ----> H $ BO $ + H # S is balanced. A different point of view: another way to write the equations for balancing the chemical reaction: #B"$ B œ ! $B"% B œ ! #B#$% $B #B œ ! B$B#$ œ! Boron Water Boric Hydrogen Sulfide Acid Sulfide Boron atoms 2 !"! Sulfur atoms 3 !!" Hydrogen atoms 0 #$# Oxygen atoms 0 "$! á #!"! $!!" B"B#$% œB B !#$# !"$! Å “ingredients” vector for boron sulfide, ...etc. or #! "!! $! ! "! B"B#$ B B % œ ! # $ # ! !" $!! compared to #B"$ B œ ! $B"% B œ ! #B#$% $B #B œ ! B$Bœ!#$ Exercise Suppose we start with the augmented matrix for a system of linear equations and using EROs we get to the matrix shown below. How many equations and variables in the original system? What can you say about solutions of the system? "!!!! $ "!!!! $ !"!!! " !"!!! " !!"!! # !!"!! # !!!"! $ !!!"! $ !!!!" ( !!!!! ( "!!"! $ !"!!! " "#!!$ !!"!! # !!!"# !!!!" ( !!!!! !.