Hi
You are right
I presume a mistake in the equations : maybe they should be
(1) 6x + 3ky = 9
(2) kx + 8y = 6
The determinant of the matrix is 48 - 3kČ, which leads to k=4 or k=-4
Posted on Friday, 14 August, 2009 - 06:24 pm:
hi..thanks
Question: Find the two values of k for which the equations
(1) 6x + 3y = 9
(2) kx + 8y = 6
do not have a unique solution. In both cases find the solution set for the equations
My attempt:
I'm looking for the values of which the coefficient matrix is zero this is 48 - 3k = 0 or k =16
I input these into the equation and obtain
(1') 6x + 3y = 9
(2') 16x + 8y = 6
or equivalently
(1') y = 3 - 2x
(2') y = (2/4)- 2x
from which I conclude that the equations are not consistent, rather they represent two parallel planes that do not meet, i.e. no solution
From here on I don't know what to do to obtain a second value of k, presumably the one where the lines are the same, i can't think that this would be true for any value of k.
Thanks, also here's the book's answer:
k=4. (3/2, -2t, t); k = -4, no solution (personally I think the book is bonkers)