Question

Let \[\begin{vmatrix}x & 2 & x \\ x^2 & x & 6 \\ x & x & 6\end{vmatrix} = a x^4 + b x^3 + c x^2 + dx + e\]
 Then, the value of \[5a + 4b + 3c + 2d + e\] is equal to

Options

• 0

• - 16

• 16

• none of these

Answer 1

 none of these
\[∆ = \begin{vmatrix} x & 2 & x\\ x^2 & x & 6\\x & x & 6 \end{vmatrix}\]
\[ = x \begin{vmatrix} x & 6\\x & 6 \end{vmatrix} - x^2 \begin{vmatrix} 2 & x\\ x & 6 \end{vmatrix} + x\begin{vmatrix} 2 & x\\ x & 6 \end{vmatrix} \left[\text{ Expanding along }C_1 \right]\]
\[ = 0 - x^2 \left( 12 - x^2 \right) + x\left( 12 - x^2 \right)\]
\[ = x^4 - 12 x^2 + 12x - x^3 \]
\[ ∆ = a x^4 + b x^3 + c x^2 + dx + e \left[\text{ Given }\right]\]
\[ \Rightarrow x^4 - 12 x^2 + 12x - x^3 = a x^4 + b x^3 + c x^2 + dx + e \]
\[ \Rightarrow a = 1, b = - 1, c = - 12, d = 12, e = 0\]
Thus, 
\[5a + 4b + 3c + 2d + e = 5 - 4 - 36 + 24 + 0 = - 11\]