# Solving system of linear equations

• April 3rd 2011, 01:19 PM
patzer
Solving system of linear equations
Find the parameter $m$ in order that the system of linear equations:

$x+y+z=2$
$2x+3y+2z=5$
$2x+3y+(2m-1)z=2m+1$

a) has one unique solution
b) has no solutions
c) has infinitely many solutions

Thank you
• April 3rd 2011, 01:46 PM
HallsofIvy
Okay, what have you done? What happens if you try to solve x, y, and z? I suspect that you will get x, y, or z, whichever you solve for first, equal to a fraction. There will be one solution if the denominator is not 0, no solution if the denominator is 0 but the numerator is not, and an infinite number of solutions if both numerator and denominator are 0 . What values of a make each of those things happen?

(Look closely at the last two equations. Tthere is an obvious first step that will simplify things very fast!)
• April 4th 2011, 02:10 PM
patzer
Hi HallsofIvy,

Using matrices, I came to an equivalent system of linear equations and found that for m=3/2 the system is impossible. For any other value other than m=3/2 the equation has an unique solution. (I don't know if this statement is completely right, and if it's an answer to the first two questions).

I don't know how to find a value of m in order that the system can have infinitely many solutions.

If you can write me the complete solution and required information, I would be really grateful.

Thank you.
• April 5th 2011, 05:45 AM
HallsofIvy
Then you have done very well.

If you subtract the second equation from the third, both x and y are eliminated and you have
$(2m- 3)z= 2m- 4$

If m 2m- 3 is NOT 0, that is, if m is NOT 3/2, then you can solve for z by dividing- if m is not 3/2, then there exist a single solution for z. If m= 3/2, the the equation becomes 0z= 3- 4= -1 which is impossible. There is no solution.

Because there is no m in either of the first two equations, if m is not 3/2, we can put z= (2m-4)/(2m-3) into the other two equations and find no new restriction on x and y. There is NO value of m for which there are an infinite number of solutions.

(If the equations had reduced to, say, (m-2)(m+1)z= m+1, then we could say that, for m= 2, we get 0z= 3 which is impossible but for m=-1, it reduces to 0z= 0 which is true for all z. That would give an infinite number of solutions but that does not happen in the given problem.)
• April 5th 2011, 07:12 AM
patzer
Thank you very much for confirmation.