For a certain value of k, the system
x + y + 3z &= 10
-4x + 3y + 5z &= 7
kx + z &= 3
has no solutions. What is this value of k?
The third equation has no y-term, so let's eliminate the y-term from the combination of the first two equations.
x + y + 3z = 10 ---> x -2 ---> -2x - 2y - 6z = -20
-4x + 3y + 5z = 7 ---> -4x + 3y + 5z = 7
Add down the columns: -6x - z = -13
Since kx + z = 3 ---> z = 3 - kx
Substituting these: -6x - (2 - 3kx) = -15
-6x - 2 + 3kx = -15
6x - 2 - 3kx = 15
6x - 3kx = 1
3x(2 - k) = 17
If 2 - k = 0, there is no value for x that will result in a product of 10,
so k = 2 results in no possible solution.
For every other value of k, there will be a value of x that can produce a product of 10.
So the answer is k = 2.
Mutiply the first equation through by -3
-3x - 3y - 9z = -30 add this to the second equation
-7x - 4z = -23
Multiply the last equation through by -4 and we have that
-4kx - 4z = -12
Note that when k = 7/4 we have the system
-7x -4z = -23
-7x - 4z = -12
This will have no solution
So....we will have no solution when