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: $abc ≥ 27$
Answer & explanation
Correct answer: option 2
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.