Practicing Success
Consider the system of equations $(a-1) x-y-z=0$ $x-(b-1) y + z=0$ $x+y-(c-1)z=0$ where a, b and c are non-zero real numbers. Statement-1: If $x, y, z$ are not all zero, then $ab + bc + ca = abc$ Statement-2: $abc ≥ 27$ |
Statement-1 is True, Statement-2 is true; Statement-2 is a correct explanation for Statement-1. Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1. Statement-1 is True, Statement-2 is False. Statement-1 is False, Statement -2 is True. |
Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for Statement-1. |
It is given that x, y, z are not all zero. So, the given system of equations has a non-trivial solutions. $∴\begin{vmatrix}a-1&-1&-1\\1&-(b-1)&1\\1&1&-(c-1)\end{vmatrix}=0$ $⇒\begin{vmatrix}a&0&-1\\1&-b&1\\c&c&1-c\end{vmatrix}=0$ [Applying $C_1→C_1-C_3,C_2 →C_2-C_3$] $⇒a (-b+bc-c) -(0 + bc) = 0$ $⇒ab + bc + ac = abc$ Using A.M. G.M., we have $⇒\frac{ab + bc + ca}{3}≥(abc)^{2/3}$ $⇒abc≥3(abc)^{2/3}⇒abc≥27$ So, both statements are true. |