Math tutor or someone whos confident with math please help !!?

Find the conditions on the numbers a, b and c that the given system has no solution, a unique solution , or infinitely many solutions.





(x1) - 2(x2) + 2(x3) = a


-2(x1) + (x2) + (x3) = b


(x1) - 5(x2) + 7(x3)= c





Please explain how u did it

Math tutor or someone whos confident with math please help !!?
[ 1 -2 2 | a]


[-2 1 1 | b]


[1 -5 7 | c]





I won't go through the details, but what we would do is reduce the matrix to row echelon form:





[1 -2 2 | a]


[0 1 -5/3 | (b+2a)/(-3)]


[0 0 0 | c - 3a - b]





In order for this system to have no solution, c - 3a - b CANNOT EQUAL 0.


If c - 3a - b does equal 0, then we have infinitely many solutions.





There is no way that the system can have a unique solution.
Reply:2(x1) - 4(x2) + 4(x3) = 2a


-2(x1) + (x2) + (x3) = b


-(x1) + 5(x2) - 7(x3)= - c


3(x2) - 5(x3) = a - c


-3(x2) + 5(x3) = 2a + b


no solution because you cannot isolate x2 and x3
Reply:Either with a matrix or Cramer's Rule, is that what it was called? hmmm...





Quite frankly, I don't remember Cramer's Rule since it has been 5 years since I took Algebra II.





So, constructing a matrix:





1 -2 2 | 1


-2 1 1 | 1


1 -5 7 | 1





Then solve the matrix (again, long forgotten and long left up to my calculator)





Alas, in the end, there is no soultion.





If you have a TI-89, just us the simultaneous equation solver. If you have a TI-83/TI-84, it can do it too, just I don't remember how.





Sorry I couldn't be more help.