Question

If $\left|\begin{array}{ccc}x& 2& x\\ x^2& x& 6\\ x& x& 6\end{array}\right|=a{x}^4+b{x}^3+c{x}^2+dx+e,$ then $5a+4b+3c+2d+e$ is equal to

• A. 11
• B. -11
• C. 12
• D. -12
• E. 13

Answer 1

Answer: (B)

Given that, $\left|\begin{array}{ccc}x& 2& x\\ x^2& x& 6\\ x& x& 6\end{array}\right|=a{x}^4+b{x}^3+c{x}^2+dx+e$
$\Rightarrow x(6x-6x)-2\left(6x^2-6x\right)+x\left(x^3-x^2\right)$
$=a{x}^4+b{x}^3+c{x}^2+dx+e$
$\Rightarrow -12x^2+12x+x^4-x^3=a{x}^4+b{x}^3+c{x}^2+dx+e$
$\Rightarrow x^4-x^3-12x^2+12x=a{x}^4+b{x}^3+c{x}^2+dx+e$
On equating the coefficient of both sides, we get
$a=1,b=-1,c=-12,d=12,e=0$
$\therefore 5a+4b+3c+2d+e=5\times 1+4\times (-1)+3(-12)$
$+2(12)+0$
$=5-4-36+24$
$=29-40=-11$