For a certain value of $k$, the system \begin{align*} x + y + 3z &= 10, \\ -4x + 2y + 5z &= 7, \\ kx + z &= 3 \end{align*} has no solutions. What is this value of $k$?
x + y + 3z = 10 multiply through by -2 ⇒ -2x - 2y - 6z = -20 (1)
-4x +2y + 5z = 7 (2)
kx + z = 3 (3)
Add (1) and (2) and we have that -6x - z = -17 multiply through by -1 ⇒ 6x + z = 17 (4)
So....we have that....using (3) qnd (4)
6x + z = 17 and
kx + z = 3
And this will have no solution if k = 6