



Now
let's look at a group of 3 equations with three variables ...






Example: 


Solve
these using matrix algebra ... 






2X + 5Y  Z = 21 



5X  Y + 4Z = 31 



2X  3Y  5Z = 31 





The
way we solve this one starts out the same, but bigger. 











The
way we combine these is a bit different. 


To
explain it, we're going to use a matrix with just letters in it. 











Now
we combine the smaller 4 number (letters here) matrices (plural of
matrix) 


the
way we did on the last page ... 





D
= A × ((E × I)  (H × F))  B × ((D × I)  (G × F)) + C × ((D
× H) 
(G × E))






Don't
even think about memorizing this equation 


just
remember how to find it. 





The
matrices with 4 numbers in them 


are
always done the same way. 





There
are many ways that the bigger 9 number matrices 


can
be worked out. 





This
one is as good as any other. 





1) 
Cross out
the top row, then cross out the left column. 



What's
left is a matrix with 4 numbers in it. 



The 4
numbers get multiplied together 



and then
multiplied by the number 



that was
crossed out twice in the big matrix. 






2) 
Cross out
the top row, then cross out the middle column. 



What's
left is a matrix with 4 numbers in it. 



This gets
multiplied together 



and then
multiplied by the number 



that was
crossed out twice in the big matrix. 



This
value will be subtracted from the other two values. 






3) 
Cross out
the top row, then cross out the right column. 



What's
left is a matrix with 4 numbers in it. 



This gets
multiplied together 



and then
multiplied by the number 



that was
crossed out twice in the big matrix. 






4) 
Add the
first and third number and subtract the second. 



THIS IS
THE ANSWER 





Let's
get back to our problem ... 











So: 



D = 2×((1×(5))((3)×4))5×((5×(5))(2×4))+(1)×((5×(3))(2×(1)) 



D = 2×((5)(12))5×((25)(8))+(1)×((15)(2)) 



D = 2×(17)5×(33)+(1)×(13) 



D = 34 + 165 + 13 



D = 212 





Now
we find Dx by substituting the answers for the X coefficients 


and
doing the same process all over again 












D_{x} = 21×((1×(5))((3)×4))(5)×((31×(5))(31×4))+(1)×((31×(3))(31×(1)) 



D_{x} = 21×((5)(12))(5)×((155)(124))+(1)×((93)(31)) 



D_{x} = 21×(17)(5)×(31)+(1)×(124) 



D_{x} = 357
+ 155 + 124 



D_{x} = 636 









Now
find D_{y} by substituting the answers for the Y
coefficients ... 












D_{Y} = 2×((31×(5))((31)×4))(21)×((5×(5))(2×4))+(1)×((5×(31))(2×(31)) 



D_{Y} = 2×((155)(124))(21)×((25)(8))+(1)×((155)(62)) 



D_{Y} = 2×(31)(21)×(33)+(1)×(217) 



D_{Y} =  62
+ 693 + 217 



D_{Y} = 848 









Now
we need to find Z. 


Once
we have X and Y, we can use substitution to find Z. 


In
fact, once we have ONE value, 


we
can use substitution to find the others. 


But
just for practice or because it's so much fun (yeah right), 


lets
use matrices to find Z ... 












D_{Z} = 2×((1×(31))((3)×31))(5)×((5×(31))(2×31))+(21)×((5×(3))(2×(1)) 



D_{Z} = 2×((31)(93))(5)×((155)(62))+(21)×((15)(2)) 



D_{Z} = 2×(124)(5)×(217)+(21)×(13) 



D_{Z} = 248
+ 1085  273 



D_{Z} = 1060 









As
you can see, 


the
chances of making simple math errors on this stuff is
really big. 


Computers
don't make that kind of mistake 


so
they love this stuff! 





copyright 2005 Bruce Kirkpatrick 
