Question

The number of values of k for which the linear equations 4x + ky + 2z = 0, kx + 4y + z = 0 and 2x + 2y + z = 0 possess a non-zero solution is

पर्याय

• 2

• 1

• zero

• 3

Answer 1

2

Explanation:

Δ = 0 ⇒ ${\displaystyle \left|\begin{array}{ccc}4& k& 2\\ k& 4& 1\\ 2& 2& 1\end{array}\right|}$ = 0

⇒ ${\displaystyle 4(4-2)-k(k-2)+2(2k-8)}$ = 0

⇒ ${\displaystyle 8-k^2+2k+4k-16}$ = 0

⇒ ${\displaystyle k^2-6k+8}$ = 0

⇒ ${\displaystyle (k-4)(k-2)}$ = 0

⇒ ${\displaystyle k}$ = 4, 2