Lesson Solving system of linear equation in 19 unknowns

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Solving system of linear equation in 19 unknowns


You can consider this problem as a  Math joke,  or as a  Math entertainment,  or seriously.
In any case,  my goal is to teach you.

Problem 1

If  a%5B1%5D, a%5B2%5D, . . ., a%5B19%5D  satisfy
    a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D = 1,        
    a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D+%2B+a%5B5%5D = 2,        
    a%5B3%5D+%2B+a%5B4%5D+%2B+a%5B5%5D+%2B+a%5B6%5D = 3,        
        . . . . . . . . .        
    a%5B16%5D+%2B+a%5B17%5D+%2B+a%5B18%5D+%2B+a%5B19%5D = 16,    
    a%5B17%5D+%2B+a%5B18%5D+%2B+a%5B19%5D+%2B+a%5B1%5D = 17,    
    a%5B18%5D+%2B+a%5B19%5D+%2B+a%5B1%5D+%2B+a%5B2%5D = 18,      
    a%5B19%5D+%2B+a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D = 19,      
find the value of  a%5B19%5D.

Solution

Let's write the system of equations in full

    a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D = 1           (1)
    a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D+%2B+a%5B5%5D = 2           (2)
    a%5B3%5D+%2B+a%5B4%5D+%2B+a%5B5%5D+%2B+a%5B6%5D = 3           (3)
        . . . . . . . . .        
    a%5B16%5D+%2B+a%5B17%5D+%2B+a%5B18%5D+%2B+a%5B19%5D = 16      (16)
    a%5B17%5D+%2B+a%5B18%5D+%2B+a%5B19%5D+%2B+a%5B1%5D = 17       (17)
    a%5B18%5D+%2B+a%5B19%5D+%2B+a%5B1%5D+%2B+a%5B2%5D = 18        (18)
    a%5B19%5D+%2B+a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D = 19         (19)

                                                          
Subtract equation (1) from equation (2):
    %28a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D+%2B+a%5B5%5D%29 - %28a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D%29 = 2 - 1
    a%5B5%5D+-+a%5B1%5D = 1
    a%5B5%5D = a%5B1%5D + 1

Subtract equation (2) from equation (3):
    %28a%5B3%5D+%2B+a%5B4%5D+%2B+a%5B5%5D+%2B+a%5B6%5D%29 - %28a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D+%2B+a%5B5%5D%29 = 3 - 2
    a%5B6%5D - a%5B2%5D = 1
    a%5B6%5D = a%5B2%5D + 1

In general, we have a%5Bn%2B4%5D = a%5Bn%5D+%2B+1.

This means:
    a%5B5%5D = a%5B1%5D + 1
    a%5B9%5D = a%5B5%5D+%2B+1 = a%5B1%5D+%2B+2
    a%5B13%5D = a%5B9%5D+%2B+1 = a%5B1%5D+%2B+3
    a%5B17%5D = a%5B13%5D+%2B+1 = a%5B1%5D+%2B+4

Similarly,
    a%5B6%5D = a%5B2%5D+%2B+1
    a%5B10%5D = a%5B6%5D+%2B+1 = a%5B2%5D+%2B+2
    a%5B14%5D = a%5B10%5D+%2B+1%7D%7D+=+%7B%7B%7Ba%5B2%5D+%2B+3
    a%5B18%5D = a%5B14%5D+%2B+1 = a%5B2%5D+%2B+4

also,
    a%5B7%5D = a%5B3%5D+%2B+1
    a%5B11%5D = a%5B7%5D+%2B+1 = a%5B3%5D+%2B+2
    a%5B15%5D = a%5B11%5D+%2B+1 = a%5B3%5D+%2B+3
    a%5B19%5D = a%5B15%5D+%2B+1 = a%5B3%5D+%2B+4

and
    a%5B8%5D = a%5B4%5D+%2B+1
    a%5B12%5D = a%5B8%5D+%2B+1 = a%5B4%5D+%2B+2
    a%5B16%5D = a%5B12%5D+%2B+1 = a%5B4%5D+%2B+3


Now let's use equations (1), (17), (18), and (19).

From equation  (1),    a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D = 1           (20)
From equation (17),    a%5B17%5D+%2B+a%5B18%5D+%2B+a%5B19%5D+%2B+a%5B1%5D = 17       (21)
From equation (18),    a%5B18%5D+%2B+a%5B19%5D+%2B+a%5B1%5D+%2B+a%5B2%5D = 18        (22)
From equation (19),    a%5B19%5D+%2B+a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D = 19         (23)


Substituting the expressions we find:

from (20):  %28a%5B1%5D+%2B+4%29+%2B+%28a%5B2%5D+%2B+4%29+%2B+%28a%5B3%5D+%2B+4%29+%2B+a%5B1%5D = 17
            2a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D+%2B+12 = 17
            2a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D = 5                            (17')

from (21):  %28a%5B2%5D+%2B+4%29+%2B+%28a%5B3%5D+%2B+4%29+%2B+a%5B1%5D+%2B+a%5B2%5D = 18
            a%5B1%5D+%2B+2a%5B2%5D+%2B+a%5B3%5D+%2B+8 = 18
            a%5B1%5D+%2B+2a%5B2%5D+%2B+a%5B3%5D = 10                           (18')

from (22):  %28a%5B3%5D+%2B+4%29+%2B+a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D = 19
            a%5B1%5D+%2B+a%5B2%5D+%2B+2a%5B3%5D+%2B+4 = 19
            a%5B1%5D+%2B+a%5B2%5D+%2B+2a%5B3%5D = 15                           (19') 


Add equations (17'), (18') and (19').  You will get

     4a%5B1%5D+%2B+4a%5B2%5D+%2B+4a%5B3%5D = 5 + 10 + 15 

     4a%5B1%5D+%2B+4a%5B2%5D+%2B+4a%5B3%5D = 30.


 Divide both sides by 4

     a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D = 7.5.      (24)


Now we are on the finish line.


From equation (17') subtract equation (24).  You will get  a%5B1%5D =  5 - 7.5 = -2.5.


From equation (18') subtract equation (24).  You will get  a%5B2%5D = 10 - 7.5 = 2.5.


From equation (19') subtract equation (24).  You will get  a%5B3%5D = 15 - 7.5 = 7.5.


From equation (23), a%5B19%5D = 19 - a%5B1%5D - a%5B2%5D - a%5B3%5D = 19 - 0 - (-2.5) - 2.5 - 7.5 =  11.5


For completeness, let's determine a-1 from the very first equation in this post

    a%5B1%5D = 1+-+%28a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D%29 = 1 - ((-2.5) + 2.5 + 7.5) = 1 - 7.5 = -6.5.


From this point, all 19 terms  a%5B1%5D, a%5B2%5D, a%5B3%5D, a%5B4%5D, a%5B5%5D, . . . , a%5B19%5D can be determined.

See this table below


    a_1    a_2    a_3    a_4    a_5    a_6    a_7    a_8    a_9    a_10   a_11    a_12    a_13    a_14    a_15    a_16    a_17    a_18    a_19  

   -2.5    2.5    7.5   -6.5   -1.5    3.5    8.5   -5.5   -0.5    4.5    9.5     -4.5    0.5     5.5     10.5    5.5     1.5     6.5     11.5


You may check that all given 19 original equations are satisfied.


Final Answer: a%5B19%5D = 11.5.


My other lessons in this site on determinants of  3x3-matrices and the Cramer's rule for solving systems of linear equations in three unknowns are
    - Determinant of a 3x3 matrix
    - Co-factoring the determinant of a 3x3 matrix
    - HOW TO solve system of linear equations in three unknowns using determinant (Cramer's rule)
    - Solving systems of linear equations in three unknowns using determinant (Cramer's rule)
    - Solving word problems by reducing to systems of linear equations in three unknowns
    - The tricks to solve some word problems with three and more unknowns using mental math
    - Solving systems of non-linear equations in three unknowns using Cramer's rule
    - Sometime two equations are enough to find three unknowns by an UNIQUE way
    - Two very different approaches to one word problem
    - Solving word problems in three unknowns by the backward method
    - Solving system of linear equation in 17 unknowns
    - OVERVIEW of LESSONS on determinants of 3x3-matrices and Cramer's rule for systems in 3 unknowns
under the current topic  Matrices, determinant, Cramer rule  of the section  Algebra-II.

My other lessons in this site on solving systems of linear equations in three unknowns are
    - Solving systems of linear equations in 3 unknowns by the Substitution method,
    - BRIEFLY on solving systems of linear equations in 3 unknowns by the Substitution method,
    - Solving systems of linear equations in 3 unknowns by the Elimination method  and
    - BRIEFLY on solving systems of linear equations in 3 unknowns by the Elimination method

Use this file/link  ALGEBRA-II - YOUR ONLINE TEXTBOOK  to navigate over all topics and lessons of the online textbook  ALGEBRA-II.


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