Posted by Dan on .
Consider the system
x2y=k1
4x+(1k^2)y=8
For what values of k does the system have
a. unique solution?
b. infinitely many solutions?
c. no solution?

Linear Algebra 
Steve,
rearranging things a bit, you have
y = x/2 + (1k)/2
y = 4x/(k^21) + 8/(1k^2)
for infinitely many solution, we need
y = mx+b
y = mx+c
where b=c
same slope:
1/2 = 4/(k^21)
k^21 = 8
k^2 = 9
k = ±3
b=c:
(1k)/2 = 8/(1k^2)
(1k)(1k^2) = 16
k=3
So, if k=3 there are infinitely many solutions.
If k = 3, there are no solutions.
For any other value of k, there is one solution. 
Linear Algebra 
Dan,
Can you perform using matrix? Thank you very much