SOLUTION: How to solve the equations for x, y, z in terms of k? And how to find the value of k for which there is no solution. x+y+z=0 kx+y-2z=-6 2y+(k+2)z=k-2 Thank you.

Question 1116543: How to solve the equations for x, y, z in terms of k?
And how to find the value of k for which there is no solution.
x+y+z=0
kx+y-2z=-6
2y+(k+2)z=k-2
Thank you.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
How to solve the equations for x, y, z in terms of k?
And how to find the value of k for which there is no solution.
x+y+z=0
kx+y-2z=-6
2y+(k+2)z=k-2
-----
Form the 3 by 3 matrix::
1...1...1
k...1..-2
0...2.(k+2)
----------------
Find the determinant value and solve the following equation::
k+2 +2k + 0 - [0 -4 + k^2+2] = 0
3k+2 +4-k^2-2 = 0
k^2-3k-4 = 0
(k-4)(k+1) = 0
k = 4 or k = -1
For those values of "k" the system has no solution.
-----
Cheers,
Stan H.
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