## limitations of cramer's rule

i usually solve 3 unknown sim equations using cramer's rule because it's systemic and fast, however today i encountered an equation that cramer's rule failed against age old tradition elimination and substitution method

-2A + B = -2
-2A - B - C = -4
2A - 3B - C = 0

usually i dont get nice interger numbers like that...
- how do i determine if a set of equations will fail to cramer's rule?
 Recognitions: Gold Member Science Advisor If the determinant of the original matrix equals 0 then it will fail Cramers rule, otherwise it should always work.....
 is there anyway to design a systematic way to calculate the 3 unknowns when cramer rule fails?

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## limitations of cramer's rule

 Quote by hanhao is there anyway to design a systematic way to calculate the 3 unknowns when cramer rule fails?
Not really.
Consider that:
x+y+z=1
2x+2y+2z=2
3x+3y+3z=3
doesn't have a unique solution

and
x+y+z=1
x+y+z=0
3x+8y+10z=52
doesn't have any solutions.

http://mathworld.wolfram.com/CramersRule.html
 Recognitions: Gold Member Science Advisor Staff Emeritus If the determinant of the right hand side is 0, the only way Cramer's rule can fail, there is no one solution: either there are NO values of A, B, C that make the equations true or there are an infinite number of solutions. In the particular system you give, it's not too hard to show that there are an infinite number of solutions and show how to get them. Subtract the second equation from the first and you get 2B+ C= 2. Add the first and last equations and you get -2B- C= -2. Since those are the same, you can pick either B or C however you like and solve for the other. In particular, C= 2- 2B. From the first equation A= 1+ B/2. Taking B to be any number at all, A= 1+ B/2, C= 2- 2B gives solutions to the equations. For example, if you take B= 2, then A= 2, C= -2 is a solution: -2(2)+ 2= -4+2= -2 -2(2)- 2-(-2)= -4- 2+ 2= -4 2(2)- 3(2)-(-2)= 4- 6+ 2= 0 Or, taking B= 6, then A= 4, C= -10 is a solution: -2(4)+ 6= -8+ 6= -2 -2(4)- 6- (-10)= -8- 6+ 10= -4 2(4)- 3(6)-(-10)= 8- 18+ 10= 0 Indeed, taking B= 4000, then A= 2001, C= -7998 is a solution! -2(2001)+ 4000= -4002+ 4000= -2 -2(2001)-(4000)-(-7998)= -2002- 4000+ 7998= -8002+ 7998= -4 2(2001)- 3(4000)-(-7998)= 4002- 12000+ 7998= 12000-12000= 0.

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