If the coefficient of $x^7\text{\hspace{0.17em}\hspace{0.17em}in\hspace{0.17em}\hspace{0.17em}}{(a{x}^2-\frac{1}{bx})}^{11}$ is equal to the coefficient of $x^{-7}$ in ${(ax-\frac{1}{b{x}^2})}^{11}$, then  $ab=$

$T_{r+1}={}^{11}{\text{C}}_r{\left(a{x}^2\right)}^{11-r}{(-\frac{1}{bx})}^r$

To get coefficient of $x^7 22-3r=7\Rightarrow r=5$ 

$\therefore T_6={}^{11}{\text{C}}_5-a^6{b}^{-5}{x}^7$

$t_{r+1}={}^{11}{\text{C}}_6{\left(ax\right)}^{11-r}{\left(\frac{-1}{b{x}^2}\right)}^r$

$={}^{11}{\text{C}}_r{(-1)}^e{a}^{11-r}{b}^{-r}{x}^{11-3r}$

To get coefficient of $x^{-7} 11-3r=-7\Rightarrow r=6$

$t_7={}^{11}{\text{C}}_6{(-1)}^6.a^5{b}^{-6}.x^{-7}$

$\text{Given\hspace{0.17em}cofficient\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}}{T}_7=\text{coefficient\hspace{0.17em}\hspace{0.17em}of\hspace{0.17em}\hspace{0.17em}}{t}_7$

$\Rightarrow -{}^{11}{\text{C}}_5\frac{a^6}{b^5}={}^{11}{\text{C}}_6\frac{a^5}{b^6}$

$\Rightarrow -a=\frac{1}{b}$

$\Rightarrow ab=-1$