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Question 24870: Hello!
Could you, please, give a hint of how to solve this system of linear equations? I'm at a loss to calculate any determinants here, which is required for most ways of solving, I guess, so I suppose there must be some "face-saver" here. :)
The last column is of the coefficients at variables to zero power - I didn't manage to put a line with this formula plotting system. Also the low lines are meant as ellipses.
Thank you!
Found 2 solutions by venugopalramana, fanks:
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! OK . NOW THE PROBLEM IS CLEAR.GOOD.BUT STILL I AM NOT CLEAR WHAT COURSE YOU ARE DOING AND HOW YOU GOT INTO THIS PROBLEM.ANY WAY LET ME SHOW A STEP BY STEP PROCEDURE AS DESIRED BY YOU TO GUIDE YOU TO THE ANSWER
LET US START WITH 3 UNKNOWNS ONLY .WE SHALL SEE SIMULTANEOUSLY HOW IT CAN BE EXPANDED TO MORE UNKNOWNS.
X1+X2+X3=1
...
1
A1X1+A2X2+A3X3 = B
.2
A1^2X1+A2^2X2+A3^2X3 = B^2
..3
NOW
.1*EQN.3 +R1*EQN.2+R2*EQN.1 GIVES US
X1(A1^2+R1A1+R2)+X2(A2^2+R1A2+R2)+X3(A3^2+R1A3+R2)=(B^2+R1B+R2)
.4
BY THIS WE BROUGHT COEFFICIENTS OF ALL UNKNOWNS AND THE CONSTANT TERM TO A UNIFORM POLYNOMIAL
LET US CALL THE POLYNOMIALS IN BRACKETS ON L.H.S. AS P2(C) INDICATING POLYNOMIAL OF DEGREE 2 OF COEFFICIENTS AND ON THE R.H.S. AS P2(K) FOR CONSTANTS.
YOU CAN EXPAND BY THIS FOR N VARIABLES... YOU WILL GET P(N-1) ( C ) AND P(N-1)(K)
NOW LET US TRY TO FIND X1 FIRST.LATER WE CAN FIND X2 AND X3 IN A SIMILAR MANNER.
SINCE WE USED THE MULTIPLYING FACTORS ..R1 AND R2 ARBITRARILY,LET US SELECT THEM SO THAT THE COEFFICIENTS OF X2 AND X3 VANISH THAT IS
A2^2+R1A2+R2 = 0
..
..4
..AND
A3^2+R1A3+R2 = 0
..5
THESE ARE HOMOGENIOUS EQNS.AND WE CAN SOLVE THEM AS FOLOWS TO FIND R1 AND R2.
SINCE BOTH ARE SAME TYPE POLYNOMIALS WE CAN TAKE THEM AS SOLUTIONS A2 AND A3 OF THE POLYNOMIAL
Y^2+R1Y+R2=0
.6
SINCE A2 AND A3 ARE SOLUTIONS OF THE ABOVE EQN.6,WE HAVE
(Y^2+R1Y+R2) - (Y-A2)(Y-A3)=0
.7
YOU CAN CHECK THIS EQN.7 FOR VALIDITY BY SUBSTITUTING Y =A2 AND Y=A3 AND USING EQNS.4 AND 5 .
SIMPLIFYING EQN.7 WE GET
Y{R1+(A2+A3)}+{R2-A2A3}=0
8
EQN.8 IS A FIRST DEGREE POLYNOMIAL IN Y AND IT HAS 2 ROOTS A2 AND A3.HENCE IT NUST BE AN IDENTITY.THAT IS
COEFFICIENTS OF EACH POWER OF Y SHALL VANISH.HENCE
R1+A2+A3=0
OR
.R1=-(A2+A3)
..LET US CALL IT - S1 TO INDICATE TAKING SUM OF THE COEFFICIENTS ONE AT A TIME.
AND
R2-A2A3 =0
OR
R2 = A2A3
..LET US CALL IT
..S2 TO INDICATE TAKING SUMOF THE COEFFICIENTS 2 AT A TIME
SO IN GENERAL CASE WHEN THERE ARE SAY A2,A3,A4,A5
FOR THE PURPOSE OF FINDING X1...THEN
S1=A2+A3+A4+A5
S2=A2A3+A2A4+A2A5+A3A4+A3A5+A4A5
S3=A2A3A4+A2A3A5+A3A4A5
S4=A2A3A4A5
..ETC
..
AND OUR MULTIPLYING COEFFICIENTS WOULD BE FOR THIS PURPOSE OF FINDING X1
.
R1=-S1
.R2=S2
..R3=-S3
.R4=S4
.ETC
HENCE WE HAVE NOW OUR VALUE OF X1 FROM EQN.4 AS
.
X1(A1^2+R1A1+R2)+X2(A2^2+R1A2+R2)+X3(A3^2+R1A3+R2)=(B^2+R1B+R2)
.4
X1(A1^2-S1A1+S2)+X2*0+X3*0=(B^2-S1B+S2)
HERE WE CAN SEE FROM EQN.7 THAT
A1^2-S1A1+S2=(A1-A2)(A1-A3)
SINCE A2 AND A3 ARE SOLUTIONS TO EQN.7 AS WE NOTED ABOVE.
X1=(B^2-S1B+S2)/(A1-A2)(A1-A3)
IN THE GENERAL CASE WE SHALL HAVE
X1={B^(N-1)-S1B^(N-2)+S2B^(N-3)-
ETC}/{(A1-A2)(A1-A3)(A1-A4)
..ETC}
NOW I THINK YOU CAN FIND X2,X3,X4 ETC
IN A SIMILAR MANNER EACH TIME CHOOSING THE MULTIPLIERS
R1,R2,R3
ETC. SUCH THAT COEFFICIENTS OF ALL UNKNOWNS EXCEPT THE ONE YOU WANT TO FIND SAY X2 ARE ZEROES.
THIS COMPLETES THE SOLUTION OF YOUR GENERAL PROBLEM TOO.HOPE YOU UNDERSTOOD.I TRIED TO MAKE THIS AS SIMPLE AS POSSIBLE.ONLY ONCE I USED THE PROPERTY OF POLYNOMIAL BECOMING AN IDENTITY..IF YOU HAVE ANY DOUBTS PLEASE WRITE BACK.
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HAVE YOU COPIED THE PROBLEM PROPERLY ?
ARE YOU SURE THE LAST ROW IS
a[1]^(n-1), a[2]^(n-1), _, a[n]^(n-1), b^(n-1)
AND NOT
a[1]^(n), a[2]^(n), _, a[n]^(n), b^(n)
AND FURTHER......
THE LAST COLUMN IS
1
B
B^2
...
...
B^N-1
OR NOT....IF SO THERE ARE N ROWS IN THIS COLUMN, WHERE AS THE FIRST COLUMN ETC HAVE ONLY N-1 ROWS.CHECK THE PROBLEM AND COME BACK
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WELL ,YOUR MATRIX IS STILL CONFUSING .ANY WAY LET ME TAKE MY OWN INTERPRETATION OF YOUR PROBLEM AND GIVE YOU THE SOLUTION.
I DO NOT UNDERSTAND YOUR FIRST ROW OF THE MATRIX.LET ME IGNORE IT.
A1+A2+A3+
...=B
A1^2+A2^2+A3^2
=B^2
A1^3+A2^3+A3^3
=B^3
.
A1^N+A2^N+A3^N
..=B^N
IF YOU KNOW THEORY OF POLYNOMIALS , I SHALL GIVE YOU THE EXPLANATION FOR THE FOLLOWING OBVIOUS ANSWER.
THIS HAS SEVERAL SOLUTION SETS AS FOLLOWS
A1=B AND A2=A3=A4=
..=0
OR
A2=B AND A1=A3=A4=
..=0
OR
A3=0 AND A1=A2=A4=
=0
ETC
..
PLEASE CONFIRM THE PROBLEM BY ELABORATING FULLY AS I GAVE ABOVE
ALSO INFORM OF YOUR BACKGROUND ON POLYNOMIALS SO THAT I CAN GIVE YOU THE PROOF IN FULL.
AS DESIRED I AM GIVING PROOF FOR THIS
PROOF OF ABOVE RESULT..
OK LET US PROCEED AS BEFORE WITH 3 UNKNOWNS AND YOU CAN GENERALISE ON THAT BASIS...
A1+A2+A3=B........................................................1
A1^2+A2^2+A3^2=B^2..............................................2
A1^3+A2^3+A3^3=B^3.................................................3
LET US TAKE EQN.1 AS BASIS
(A1+A2+A3)^2=B^2
A1^2+A2^2+A3^2+2A1A2+2A1A3+2A2A3=B^2
B^2+2(A1A2+A1A3+A2A3)=B^2
..USING EQN.2
(A1A2+A1A3+A2A3)=0
.4
SIMILARLY
(A1+A2+A3)^3=B^3
A1^3+A2^3+A3^3+3(A1+A2+A3)(A1A2+A1A3+A2A3)-3A1A2A3=B^3
.USING WELL KNOWN EXPANSION OF (A+B+C)^3
(A1+A2+A3)(A1A2+A1A3+A2A3) - A1A2A3=0
SUBSTITUTING EQN.4 IN THIS ,
.WE GET
A1A2A3 = 0
.5
FROM EQN.5 ,WE HAVE A1=0
OR
A2=0
OR
A3=0
TAKING A1=0 SAY
FROM EQN.4
WE GET
..A2A3=0
.HENCE EITHER A2=0 OR A3=0
.TAKING A2=0 SAY..
FROM EQN.1 WE GET 0+0+A3=B..OR
A3=B
.HENCE ONE SOLUTION SET IS
A1=0 AND A2=0 AND A3=B
OTHERS FOLLOW IN A SIMILAR MANNER..YOU CAN EXPAND LIKE WISE FOR 4 UNKNOWNS BY FORMING
A 4TH DEGREE POLYNOMIAL LIKE
Y^4-PY^3+QY^2-RY+T=0
AND USE OUR NOMENCLATURE LIKE
A1^K+A2^K+A3^K+A4^K=SK
.HENCE
S4-PS3+QS2-RS1+T=0
WHERE
S4=B^4,S3=B^3,S2=B^2,S1=B
ETC
NOTE THAT THIS EQN. HAS THE ROOT Y=B AND HENCE LET A1=B
AND HENCE
THE OTHERS WILL BECOME ZERO FROM THE PROOF WE GAVE FOR 3 UNKNOWNS
.ETC
Answer by fanks(6)  (Show Source):
You can put this solution on YOUR website! Thanks for the response! :)
Yes, sorry, there is another row at the beginning. Without b's it's n x n matrix. I edited the question formula accordingly.
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