Question

If $\Delta \left(x\right)=\left|\begin{array}{ccc}x-2& (x-1{)}^2& x^3\\ x-1& x^2& (x+1{)}^3\\ x& (x+1{)}^2& (x+2{)}^3\end{array}\right|$, find the absolute value of coefficient of $x$ in $\Delta \left(x\right)$

Answer 1

$\Delta \left(x\right)=\left|\begin{array}{ccc}x-2& (x-1{)}^2& x^3\\ x-1& x^2& (x+1{)}^3\\ x& (x+1{)}^2& (x+2{)}^3\end{array}\right|$

$\Delta \left(x\right)$ polynomial of degree atmost $6$ in $x$.

If $\Delta \left(x\right)=a_0+a_{1}x+a_2{x}^2+...+a_6{x}^6$, then

${\Delta }^{\prime }\left(x\right)=a_1+2a_{2}x+...+6a_6{x}^5\Rightarrow a_1={\Delta }^{\prime }\left(0\right)$

${\Delta }^{\prime }\left(x\right)=\left|\begin{array}{ccc}1& {(x-1)}^2& x^3\\ 1& x^2& {(x+1)}^3\\ 1& {(x+1)}^2& {(x+2)}^3\end{array}\right|+\left|\begin{array}{ccc}x-2& 2(x-1)& x^3\\ x-1& 2x& (x+1{)}^3\\ x& 2(x+1)& (x+2{)}^3\end{array}\right|+\left|\begin{array}{ccc}x-2& (x-1{)}^2& 3x^2\\ x-1& x^2& 3(x+1{)}^2\\ x& (x+1{)}^2& 3(x+2{)}^2\end{array}\right|$

${\Delta }^{\prime }\left(0\right)$$=\left|\begin{array}{ccc}1& 1& 0\\ 1& 0& 1\\ 1& 1& 8\end{array}\right|+\left|\begin{array}{ccc}-2& -2& 0\\ -1& 0& 1\\ 0& 2& 8\end{array}\right|+\left|\begin{array}{ccc}-2& 1& 0\\ -1& 0& 3\\ 0& 1& 12\end{array}\right|$

$=-8-12+18=-2$

$\therefore$ Absolute value of coefficient of $x$ is $2$