Question

Let $\left|\begin{array}{ccc}x& 2& x\\ x^2& x& 6\\ x& x& 6\end{array}\right|=A{x}^4+B{x}^3+C{x}^2+Dx+E$. then the value of $5A+4B+3C+2D+E$ is equal to

• A. zero
• B. $-16$
• C. $16$
• D. $-11$

Answer 1

Answer: (D)

$-11$

Let the given determinant be equal to $\Delta \left(x\right)$. Then

$5A+4B+3C+2D+E=\Delta \left(1\right)+{\Delta }^{\prime }\left(1\right)$

now $\Delta \left(1\right)$ as $R_2$ and $R_3$ are identical $\Rightarrow \Delta \left(1\right)=0$

${\Delta }^{\prime }\left(x\right)=\left|\begin{array}{ccc}1& 0& 1\\ x^2& x& 6\\ x& x& 6\end{array}\right|+\left|\begin{array}{ccc}x& 2& x\\ 2x& 1& 0\\ x& x& 6\end{array}\right|+\left|\begin{array}{ccc}x& 2& x\\ x^2& x& 6\\ 1& 1& 6\end{array}\right|;{\Delta }^{\prime }\left(1\right)=\left|\begin{array}{ccc}1& 2& 1\\ 2& 1& 0\\ 1& 1& 6\end{array}\right|+\left|\begin{array}{ccc}1& 2& 1\\ 1& 1& 6\\ 1& 1& 0\end{array}\right|=-17+(12+1-1-6)=-11$

$\therefore 5A+4B+3C+2D+E=\Delta \left(1\right)+{\Delta }^{\prime }\left(1\right)=0+(-11)=-11$