SOLUTION: show that if Ax=b has more than one solution, then it has infinitely many solutions(Hint: If x1 and x2 are solutions, consider x3=rx1 +sx2, where r+s=1)

Algebra ->  Matrices-and-determiminant -> SOLUTION: show that if Ax=b has more than one solution, then it has infinitely many solutions(Hint: If x1 and x2 are solutions, consider x3=rx1 +sx2, where r+s=1)      Log On


   



Question 25009: show that if Ax=b has more than one solution, then it has infinitely many solutions(Hint: If x1 and x2 are solutions, consider x3=rx1 +sx2, where r+s=1)
Answer by venugopalramana(3286) About Me  (Show Source):
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show that if Ax=b has more than one solution, then it has infinitely many solutions(Hint: If x1 and x2 are solutions, consider x3=rx1 +sx2, where r+s=1)
AX=B
LET THE 2 SOLUTIONS BE X1 AND X2.HENCE
AX1=B.....................I... AND
AX2=B.....................II
MULTIPLYING X1 WITH A CONSTANT r,WE GET FROM EQN.I
A*rX1=r*AX1=rB.............III
SIMILARLY MULTIPLYING X2 WITH A CONSTANT s,WE GET FROM EQN.II
A*sX2=s*AX2=sB.............IV
ADDING EQNS.III AND IV,WE GET
A*(rX1+sX2)=(r+s)B
IF WE TAKE THE 2 CONSTANTS r AND s SUCH THAT r+s=1,WE HAVE
A*(rX1+sX2)=1B=B
HENCE (rX1+sX2) IS ALSO A SOLUTION.THUS BY TAKING DIFFERENT VALUES FOR r AND SUCH THAT r+s=1,WE GET INFINITE SOLUTIONS.