If the coefficient of x⁷ in the expansion of  ${\left(a{x}^2+\frac{1}{bx}\right)}^{11}$ and the coefficient of x^{– 7} is the expansion of ${\left(ax-\frac{1}{b{x}^2}\right)}^{11}$are equal, then

$T_{r+1},$the${(r+1)}^{\text{ th }}$ term in the expansion ${\left(a{x}^2+\frac{1}{bx}\right)}^{11}$is 

$T_{r+1}{=}^{11}{C}_r{\left(a{x}^2\right)}^{11-r}{\left(\frac{1}{bx}\right)}^r{=}^{11}{C}_r\frac{a^{11-r}}{b^r}{x}^{22-3r}$

For the coefficient of $x^7,$set $22-3r=7\Rightarrow r=5$

$\therefore$Coefficient of x⁷ in ${\left(a{x}^2+\frac{1}{bx}\right)}^{11}$is ${ }^{11}{C}_5\frac{a^6}{b^5}$

Next, $t_{r }+ 1$, the ${(r + 1)}^{th}$ term in the expansion of ${\left(ax-\frac{1}{b{x}^2}\right)}^{11}$is 

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

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

For the coefficient of $x^{-7},$ set $11-3r=-7\Rightarrow r=6$

$\therefore$Coefficient of $x^{-7}$ in the expansion of ${=}^{11}{C}_6\frac{a^5}{b^6}$

We are given  ${ }^{11}{C}_5\frac{a^6}{b^5}{=}^{11}{C}_6\frac{a^5}{b^6}\Rightarrow ab=1$