Question

If $f\left(x\right)=\left|\begin{array}{ccc}(1+x{)}^{17}& (1+x{)}^{19}& (1+x{)}^{23}\\ (1+x{)}^{23}& (1+x{)}^{29}& (1+x{)}^{34}\\ (1+x{)}^{41}& (1+x{)}^{43}& (1+x{)}^{47}\end{array}\right|=A+Bx+C{x}^2......,$ then A = .......

Answer 1

$f\left(x\right)\left|\begin{array}{ccc}(1+x{)}^{17}& (1+x{)}^{19}& (1+x{)}^{23}\\ (1+x{)}^{23}& (1+x{)}^{29}& (1+x{)}^{34}\\ (1+x{)}^{41}& (1+x{)}^{43}& (1+x{)}^{47}\end{array}\right|=A+Bx+(x^2+...)$
Consider $\left|\begin{array}{ccc}(1+x{)}^{17}& (1+x{)}^{19}& (1+x{)}^{23}\\ (1+x{)}^{23}& (1+x{)}^{29}& (1+x{)}^{34}\\ (1+x{)}^{41}& (1+x{)}^{43}& (1+x{)}^{47}\end{array}\right|$
$=(1+x{)}^{17}(1+x{)}^{23}(1+x{)}^{41}\left|\begin{array}{ccc}1& (1+x{)}^2& (1+x{)}^6\\ 1& (1+x{)}^6& (1+x{)}^{11}\\ 1& (1+x{)}^2& (1+x{)}^6\end{array}\right|$
Row 1 = Row 3
$\Rightarrow \left|\begin{array}{ccc}1& (1+x{)}^2& (1+x{)}^6\\ 1& (1+x{)}^6& (1+x{)}^{11}\\ 1& (1+x{)}^2& (1+x{)}^6\end{array}\right|=0$
$\Rightarrow (1+x{)}^{17}(1+x{)}^{23}(1+x{)}^{41}.0=A+Bx+C{x}^2+...$
Comparing we get A = 0.