Question

Consider the system
x-2y=k-1
4x+(1-k^2)y=8

For what values of k does the system have
a. unique solution?
b. infinitely many solutions?
c. no solution?

Answer 1

rearranging things a bit, you have

y = x/2 + (1-k)/2
y = 4x/(k^2-1) + 8/(1-k^2)

for infinitely many solution, we need
y = mx+b
y = mx+c
where b=c

same slope:
1/2 = 4/(k^2-1)
k^2-1 = 8
k^2 = 9
k = ±3

b=c:
(1-k)/2 = 8/(1-k^2)
(1-k)(1-k^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.

Answer 2

Can you perform using matrix? Thank you very much