The system of equations kx + (k + 1)y + (k - 1) z = 0 (k + 1)x + ky + (k + 2) z = 0 (k - 1)x + (k + 2)y + kz = 0 has a non-trivial solution for

Question

The system of equations kx + (k + 1)y + (k - 1) z = 0 (k + 1)x + ky + (k + 2) z = 0 (k - 1)x + (k + 2)y + kz = 0 has a non-trivial solution for (A) exactly three real values of k. (B) exactly two real values of k. (C) exactly one real value of k. (D) infinite number of values of k.

Tardigrade Admin · Accepted Answer

To have a non-trivial solution, we must have |k k+1 k-1 k+1 k k+2 k-1 k+2 k|=0 ⇒ 2 k+1=0 ⇒ k=(-1/2)

Q. The system of equations
$k x + (k + 1) y + (k - 1) z = 0$
$(k + 1) x + k y + (k + 2) z = 0$
$(k - 1) x + (k + 2) y + k z = 0$
has a non-trivial solution for

exactly three real values of k.

exactly two real values of k.

exactly one real value of k.

infinite number of values of k.

To have a non-trivial solution, we must have
$∣ ∣ k k + 1 k - 1 k + 1 k k + 2 k - 1 k + 2 k ∣ ∣ = 0 \Rightarrow 2 k + 1 = 0 \Rightarrow k = \frac{- 1}{2}$

Q. The system of equations kx+(k+1)y+(k−1)z=0 (k+1)x+ky+(k+2)z=0 (k−1)x+(k+2)y+kz=0 has a non-trivial solution for