For the given set of equations: 4x+ky+z=0 kx+4y+z=0 2x+2y+z=0 It can be observed that x = y = z = 0 is one solution of the system. In order to have infinitely many (including non-zero) solutions, D must be zero. ⇒D=|
4
k
2
k
4
2
1
1
1
|=0 ⇒4(4−2)+k(2−k)+2(k−4)=0 ⇒8+2k−k2+2k−8=0 ⇒k2+4k=0 ⇒k(k+4)=0 ⇒k=0ORk+4=0 ⇒k=0ORk=−4 Hence, there are 2 possible values of k but non-zero solution is one.