For what value(s) of $\displaystyle m$ will the system of equations,

$\displaystyle x+my=2$

$\displaystyle (m-1)x+2y=m$

have:

a) a unique solution?

b) no solution?

c) an infinite number of solutions?

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- Apr 18th 2010, 07:59 AMshawlisimultaneous equations in matrices - 3 solutions
For what value(s) of $\displaystyle m$ will the system of equations,

$\displaystyle x+my=2$

$\displaystyle (m-1)x+2y=m$

have:

a) a unique solution?

b) no solution?

c) an infinite number of solutions? - Apr 18th 2010, 08:09 AMHallsofIvy
Try solving it! For example, you can multiply the first equation by 2 and the second equation by m and subtract to eliminate y:

$\displaystyle (2x+ 2my)- (m(m-1)x+ 2my)= 4- m^2$

$\displaystyle (2- m^2+ m)x= 4- m^2$

$\displaystyle x= \frac{m^2- 4}{m^2- m- 2}$

Now, if the denominator is not 0, there is a unique solution. If the denominator is 0 and the numerator is not, there is no solution. If both numerator and denominator are 0, there are an infinite number of solutions.