Posted by **Dan** on Tuesday, December 18, 2012 at 2:42pm.

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?

- Linear Algebra -
**Steve**, Tuesday, December 18, 2012 at 2:59pm
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.

- Linear Algebra -
**Dan**, Tuesday, December 18, 2012 at 5:40pm
Can you perform using matrix? Thank you very much

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