Question

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

Answer 1

Given:

$f\left(x\right)=\left|\begin{array}{ccc}(1+x{)}^{17}& (1+x{)}^{19}& (1+x{)}^{23}\\ (1+x{)}^{23}& (1+x{)}^{29}& (1+x{)}^{34}\\ (1+x{)}^{41}& (1+x{)}^{43}& (1+x{)}^{47}\end{array}\right|$

Taking $(1+x{)}^{17},(1+x{)}^{23}$ and $(1+x{)}^{41}$ common from $R_1,R_2$ and $R_3$ respectively.

$\Rightarrow f\left(x\right)=(1+x{)}^{17}(1+x{)}^{23}(1+x{)}^{41}$

$\left|\begin{array}{ccc}1& (1+x{)}^2& (1+x{)}^6\\ 1& (1+x{)}^6& (1+x{)}^{11}\\ 1& (1+x{)}^2& (1+x{)}^6\end{array}\right|$

If any two rows or columns are identical then value of determinant is zero. Here, $R_1$ and $R_3$ are identical.

$\Rightarrow f\left(x\right)=0$

$\therefore A+Bx+C{x}^2+\dots =0$

$(\because f(x)=A+Bx+C{x}^2+\dots )$

By comparing the like terms, we get $A=0$

Hence, the value of 𝐴 is 0.