Question

if $a\ne 6,b,c$ satisfy $\left|\begin{array}{ccc}a& 2b& 2c\\ 3& b& c\\ 4& a& b\end{array}\right|=0$, then $abc=$

• A. $a+b+c$
• B. $0$
• C. $b^3$
• D. $ab+b-c$

Answer 1

The correct option is C $b^3$

Let$A=\left|\begin{array}{ccc}a& 2b& 2c\\ 3& b& c\\ 4& a& b\end{array}\right|$ and $a\ne 6$

So, By operation of matrix ,

$A=a(b^2-ac)-2b(3b-4c)+2c(3a-4b)$

$=a{b}^2-a^{2}c-6b^2+8bc+6ac-8bc$

$=6ac+a{b}^2-a^{2}c-6b^2$

Given,$A=0=6ac+a{b}^2-a^{2}c-6b^2$

$6ac+a{b}^2-a^{2}c-6b^2=0$

$ac(6-a)+b^2(a-6)=0$

$(a-6)(b^2-ac)=0$

$\therefore b^2=ac$ ( $\because a\ne 6$)

$\therefore abc=b^3$