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. -17
• B. -14
• C. -11
• D. -8

Answer 1

The correct option is C -11

Let $\Delta \left(X\right)=\left|\begin{array}{ccc}x& 2& x\\ x^2& x& 6\\ x& x& 6\end{array}\right|\Rightarrow \Delta \left(1\right)=0\therefore {\Delta }^{\prime }\left(x\right)=\left|\begin{array}{ccc}1& 0& 1\\ x^2& x& 6\\ x& x& 6\end{array}\right|+\left|\begin{array}{ccc}{x}^2& 2& x\\ 2x& 1& 0\\ x& x& 6\end{array}\right|+\left|\begin{array}{ccc}x& 2& x\\ x^2& x& 6\\ 1& 1& 0\end{array}\right|{\Delta }^{\prime }\left(1\right)=0+\left|\begin{array}{ccc}1& 2& 1\\ 2& 1& 0\\ 1& 1& 6\end{array}\right|+\left|\begin{array}{ccc}1& 2& 1\\ 1& 1& 6\\ 1& 1& 0\end{array}\right|$
=-17 +6 =-11
Now, $\Delta \left(1\right)+{\Delta }^{\prime }\left(1\right)=5A+4B+3C+4D+E=-11$
Alternative method
Multiplying both sides by x, then
$\left|\begin{array}{ccc}{x}^2& 2x& x^2\\ x^2& x& 6\\ x& x& 6\end{array}\right|=A{x}^5+B{x}^4+C{x}^3+D{x}^2+Ex$
Differentiating both sides w. r. t. x, then
$\left|\begin{array}{ccc}2x& 2& 2x\\ x^2& x& 6\\ x& x& 6\end{array}\right|+\left|\begin{array}{ccc}{x}^2& 2x& x^2\\ 2x& 1& 0\\ x& x& 6\end{array}\right|+\left|\begin{array}{ccc}{x}^2& 2x& x^2\\ x^2& x& 6\\ 1& 1& 0\end{array}\right|$
$=5A{x}^4+4B{x}^3+3C{x}^2+2Dx+E$
$\left|\begin{array}{ccc}2& 2& 2\\ 1& 1& 6\\ 1& 1& 6\end{array}\right|+\left|\begin{array}{ccc}1& 2& 1\\ 2& 1& 0\\ 1& 1& 6\end{array}\right|+\left|\begin{array}{ccc}1& 2& 1\\ 1& 1& 6\\ 1& 1& 0\end{array}\right|=5A+4B+3C+2D+E$
$\Rightarrow 0+1\left(6\right)-2\left(12\right)+1(2-1)+1(-6)-2(0-6)+1(1-1)=5A+4B+3C+2D+E$
Hence, 5A, + 4B + 3C + 2D + E = -11