Question

If $a \neq 6,b,c$satisfy $\left| \begin{matrix} a & 2b & 2c \\ 3 & b & c \\ 4 & a & b \end{matrix} \right| = 0,$ then $abc =$

• A. $a + b + c$
• B. 0
• C. $b^{3}$
• D. $ab + bc$

Answer 1

Answer: (C)

$b^{3}$

Answer 2

$\left| \begin{array} { c c c } a & 2 b & 2 c \\ 3 & b & c \\ 4 & a & b \end{array} \right| = 0 \Rightarrow \left| \begin{array} { c c c } a - 6 & 0 & 0 \\ 3 & b & c \\ 4 & a & b \end{array} \right| = 0$ $A^{3} = I$

⇒ $(a - 6)(b^{2} - ac) = 0 \Rightarrow b^{2} - ac = 0$, $= 3\left\lbrack \frac{- 1 + \sqrt{3}i}{2} - \frac{- 1 - \sqrt{3}i}{2} \right\rbrack = 3\sqrt{3}i$

$\therefore$ $ac = b^{2} \Rightarrow abc = b^{3}.$