Lesson Solving system of linear equation in 17 unknowns

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Solving system of linear equation in 17 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%5B17%5D)  satisfy
    a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D = 1, 
    a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D = 2,
    a%5B3%5D+%2B+a%5B4%5D+%2B+a%5B5%5D = 3, 
        . . . . . . . . .        
    a%5B15%5D+%2B+a%5B16%5D+%2B+a%5B17%5D = 15, 
    a%5B16%5D+%2B+a%5B17%5D+%2B+a%5B1%5D = 16, 
    a%5B17%5D+%2B+a%5B1%5D+%2B+a%5B2%5D = 17,
find the value of  a%5B17%5D.

Solution 

Let's write the system of equations in full

    a%5B1%5D+%2B+a%5B2%5D+%2B+a%5B3%5D = 1,        (1)
    a%5B2%5D+%2B+a%5B3%5D+%2B+a%5B4%5D = 2,        (2)
    a%5B3%5D+%2B+a%5B4%5D+%2B+a%5B5%5D = 3,        (3) 
    a%5B4%5D+%2B+a%5B5%5D+%2B+a%5B6%5D = 4,        (4) 
    a%5B5%5D+%2B+a%5B6%5D+%2B+a%5B7%5D = 5,        (5)
    a%5B6%5D+%2B+a%5B7%5D+%2B+a%5B8%5D = 6,        (6)
    a%5B7%5D+%2B+a%5B8%5D+%2B+a%5B9%5D = 7,        (7) 
    a%5B8%5D+%2B+a%5B9%5D+%2B+a%5B10%5D = 8,       (8) 
    a%5B9%5D+%2B+a%5B10%5D+%2B+a%5B11%5D = 9,      (9)
    a%5B10%5D+%2B+a%5B11%5D+%2B+a%5B12%5D = 10,    (10) 
    a%5B11%5D+%2B+a%5B12%5D+%2B+a%5B13%5D = 11,    (11) 
    a%5B12%5D+%2B+a%5B13%5D+%2B+a%5B14%5D = 12,    (12) 
    a%5B13%5D+%2B+a%5B14%5D+%2B+a%5B15%5D = 13,    (13)
    a%5B14%5D+%2B+a%5B15%5D+%2B+a%5B16%5D = 14,    (14)
    a%5B15%5D+%2B+a%5B16%5D+%2B+a%5B17%5D = 15,    (15)
    a%5B16%5D+%2B+a%5B17%5D+%2B+a%5B1%5D = 16,     (16)
    a%5B17%5D+%2B+a%5B1%5D+%2B+a%5B2%5D = 17.      (17)


Add all 17 equations (1) - (17)  (both sides). You will get

     = 1 + 2 + 3 + . . . + 15 + 16 + 17 = %28%281+%2B+17%29%2F2%29%2A17 = 9*17 = 153.


Now divide both sides of the lest equation by 3.  You will get

   = 51.   (18)


Next subtract equation (1), (4), (7), (10), (13) from equation (18). You will get

    a%5B16%5D%2Ba%5B17%5D = 51 - 1 - 4 - 7 - 10 - 13 = 16.   (19)


Now compare equations (19) and (16). You instantly will get 

    a%5B1%5D = 0.                      (20)


Having known a%5B1%5D = 0, we can rewrite the equation (20) in the form

     = 51.   (21)

        Now we are on the finish line, finally !!! 

Next subtract equation (2), (5), (8), (11) from equation (21). You will get

    a%5B16%5D%2Ba%5B17%5D = 51 - 2 - 5 - 8 - 11 = 25.   (22)


As the last step, subtract equation (14) from equation (22). You will get

    a%5B17%5D = 25 - 14 = 11.   (30)


It is your answer:  a%5B17%5D = 11.


For completeness, moving from equation (1) to equation (17), we can determine all 17 unknowns:

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 

     0      6     -5      1      7     -4      2      8     -3       3     9      -2        4      10     -1        5      11


You may check that all given 17 original equations are satisfied.

At this point, the problem is fully solved.

Answer.   a%5B17%5D = 11.


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 19 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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