Algebra.Com's Answer #13841 by venugopalramana(3286)![\"\"](\"/images/stars/stars-13.gif\")  
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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
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document.write( "LET US START WITH 3 UNKNOWNS ONLY .WE SHALL SEE SIMULTANEOUSLY HOW IT CAN BE EXPANDED TO MORE UNKNOWNS.\r
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document.write( "X1+X2+X3=1
...
1
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document.write( "A1X1+A2X2+A3X3 = B
.2
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document.write( "A1^2X1+A2^2X2+A3^2X3 = B^2
..3
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document.write( "NOW
.1*EQN.3 +R1*EQN.2+R2*EQN.1 GIVES US
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document.write( "X1(A1^2+R1A1+R2)+X2(A2^2+R1A2+R2)+X3(A3^2+R1A3+R2)=(B^2+R1B+R2)
.4\r
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document.write( "BY THIS WE BROUGHT COEFFICIENTS OF ALL UNKNOWNS AND THE CONSTANT TERM TO A UNIFORM POLYNOMIAL
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document.write( "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.
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document.write( "YOU CAN EXPAND BY THIS FOR N VARIABLES... YOU WILL GET P(N-1) ( C ) AND P(N-1)(K)\r
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document.write( "NOW LET US TRY TO FIND X1 FIRST.LATER WE CAN FIND X2 AND X3 IN A SIMILAR MANNER.
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document.write( "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\r
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document.write( "A2^2+R1A2+R2 = 0
..
..4
..AND
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document.write( "A3^2+R1A3+R2 = 0
..5
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document.write( "THESE ARE HOMOGENIOUS EQNS.AND WE CAN SOLVE THEM AS FOLOWS TO FIND R1 AND R2.
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document.write( "SINCE BOTH ARE SAME TYPE POLYNOMIALS WE CAN TAKE THEM AS SOLUTIONS A2 AND A3 OF THE POLYNOMIAL
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document.write( "Y^2+R1Y+R2=0
.6
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document.write( "SINCE A2 AND A3 ARE SOLUTIONS OF THE ABOVE EQN.6,WE HAVE
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document.write( "(Y^2+R1Y+R2) - (Y-A2)(Y-A3)=0
.7
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document.write( "YOU CAN CHECK THIS EQN.7 FOR VALIDITY BY SUBSTITUTING Y =A2 AND Y=A3 AND USING EQNS.4 AND 5 .
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document.write( "SIMPLIFYING EQN.7 WE GET
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document.write( "Y{R1+(A2+A3)}+{R2-A2A3}=0
8
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document.write( "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
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document.write( "COEFFICIENTS OF EACH POWER OF Y SHALL VANISH.HENCE
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document.write( "R1+A2+A3=0
OR
.R1=-(A2+A3)
..LET US CALL IT - S1 TO INDICATE TAKING SUM OF THE COEFFICIENTS ONE AT A TIME.
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document.write( "AND
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document.write( "R2-A2A3 =0
OR
R2 = A2A3
..LET US CALL IT
..S2 TO INDICATE TAKING SUMOF THE COEFFICIENTS 2 AT A TIME
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document.write( "SO IN GENERAL CASE WHEN THERE ARE SAY A2,A3,A4,A5
FOR THE PURPOSE OF FINDING X1...THEN
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document.write( "S1=A2+A3+A4+A5
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document.write( "S2=A2A3+A2A4+A2A5+A3A4+A3A5+A4A5
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document.write( "S3=A2A3A4+A2A3A5+A3A4A5
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document.write( "S4=A2A3A4A5
..ETC
..
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document.write( "AND OUR MULTIPLYING COEFFICIENTS WOULD BE FOR THIS PURPOSE OF FINDING X1
.
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document.write( "R1=-S1
.R2=S2
..R3=-S3
.R4=S4
.ETC
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document.write( "HENCE WE HAVE NOW OUR VALUE OF X1 FROM EQN.4 AS
.
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document.write( "X1(A1^2+R1A1+R2)+X2(A2^2+R1A2+R2)+X3(A3^2+R1A3+R2)=(B^2+R1B+R2)
.4
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document.write( "X1(A1^2-S1A1+S2)+X2*0+X3*0=(B^2-S1B+S2)
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document.write( "HERE WE CAN SEE FROM EQN.7 THAT
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document.write( "A1^2-S1A1+S2=(A1-A2)(A1-A3)
SINCE A2 AND A3 ARE SOLUTIONS TO EQN.7 AS WE NOTED ABOVE.\r
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document.write( "X1=(B^2-S1B+S2)/(A1-A2)(A1-A3)\r
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document.write( "IN THE GENERAL CASE WE SHALL HAVE \r
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document.write( "X1={B^(N-1)-S1B^(N-2)+S2B^(N-3)-
ETC}/{(A1-A2)(A1-A3)(A1-A4)
..ETC}\r
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document.write( "NOW I THINK YOU CAN FIND X2,X3,X4 ETC
IN A SIMILAR MANNER EACH TIME CHOOSING THE MULTIPLIERS
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document.write( "R1,R2,R3
ETC. SUCH THAT COEFFICIENTS OF ALL UNKNOWNS EXCEPT THE ONE YOU WANT TO FIND SAY X2 ARE ZEROES.
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document.write( "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.\r
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document.write( "HAVE YOU COPIED THE PROBLEM PROPERLY ?
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document.write( "ARE YOU SURE THE LAST ROW IS
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document.write( "a[1]^(n-1), a[2]^(n-1), _, a[n]^(n-1), b^(n-1)\r
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document.write( "AND NOT
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document.write( "a[1]^(n), a[2]^(n), _, a[n]^(n), b^(n)
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document.write( "AND FURTHER......
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document.write( "THE LAST COLUMN IS
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document.write( "1
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document.write( "B
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document.write( "B^2
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document.write( "...
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document.write( "...
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document.write( "B^N-1
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document.write( "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\r
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document.write( "WELL ,YOUR MATRIX IS STILL CONFUSING .ANY WAY LET ME TAKE MY OWN INTERPRETATION OF YOUR PROBLEM AND GIVE YOU THE SOLUTION.
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document.write( "I DO NOT UNDERSTAND YOUR FIRST ROW OF THE MATRIX.LET ME IGNORE IT.
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document.write( "A1+A2+A3+
...=B
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document.write( "A1^2+A2^2+A3^2
=B^2
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document.write( "A1^3+A2^3+A3^3
=B^3
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document.write( "
.
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document.write( "A1^N+A2^N+A3^N
..=B^N \r
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document.write( "IF YOU KNOW THEORY OF POLYNOMIALS , I SHALL GIVE YOU THE EXPLANATION FOR THE FOLLOWING OBVIOUS ANSWER.
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document.write( "THIS HAS SEVERAL SOLUTION SETS AS FOLLOWS
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document.write( "A1=B AND A2=A3=A4=
..=0
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document.write( "OR
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document.write( "A2=B AND A1=A3=A4=
..=0
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document.write( "OR
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document.write( "A3=0 AND A1=A2=A4=
=0
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document.write( "ETC
..
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document.write( "PLEASE CONFIRM THE PROBLEM BY ELABORATING FULLY AS I GAVE ABOVE
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document.write( "ALSO INFORM OF YOUR BACKGROUND ON POLYNOMIALS SO THAT I CAN GIVE YOU THE PROOF IN FULL.
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document.write( "AS DESIRED I AM GIVING PROOF FOR THIS
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document.write( "PROOF OF ABOVE RESULT..
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document.write( "OK LET US PROCEED AS BEFORE WITH 3 UNKNOWNS AND YOU CAN GENERALISE ON THAT BASIS...
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document.write( "A1+A2+A3=B........................................................1
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document.write( "A1^2+A2^2+A3^2=B^2..............................................2
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document.write( "A1^3+A2^3+A3^3=B^3.................................................3
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document.write( "LET US TAKE EQN.1 AS BASIS
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document.write( "(A1+A2+A3)^2=B^2
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document.write( "A1^2+A2^2+A3^2+2A1A2+2A1A3+2A2A3=B^2
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document.write( "B^2+2(A1A2+A1A3+A2A3)=B^2
..USING EQN.2
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document.write( "(A1A2+A1A3+A2A3)=0
.4
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document.write( "SIMILARLY
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document.write( "(A1+A2+A3)^3=B^3
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document.write( "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
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document.write( "(A1+A2+A3)(A1A2+A1A3+A2A3) - A1A2A3=0
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document.write( "SUBSTITUTING EQN.4 IN THIS ,
.WE GET
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document.write( "A1A2A3 = 0
.5
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document.write( "FROM EQN.5 ,WE HAVE A1=0
OR
A2=0
OR
A3=0
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document.write( "TAKING A1=0 SAY
FROM EQN.4
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document.write( "WE GET
..A2A3=0
.HENCE EITHER A2=0 OR A3=0
.TAKING A2=0 SAY..
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document.write( "FROM EQN.1 WE GET 0+0+A3=B..OR
A3=B
.HENCE ONE SOLUTION SET IS
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document.write( "A1=0 AND A2=0 AND A3=B
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document.write( "OTHERS FOLLOW IN A SIMILAR MANNER..YOU CAN EXPAND LIKE WISE FOR 4 UNKNOWNS BY FORMING
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document.write( "A 4TH DEGREE POLYNOMIAL LIKE
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document.write( "Y^4-PY^3+QY^2-RY+T=0
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document.write( "AND USE OUR NOMENCLATURE LIKE
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document.write( "A1^K+A2^K+A3^K+A4^K=SK
.HENCE
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document.write( "S4-PS3+QS2-RS1+T=0
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document.write( "WHERE
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document.write( "S4=B^4,S3=B^3,S2=B^2,S1=B
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document.write( "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
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