An algebra problem by naitik sanghavi

Since the functions $1(x)= x$ , $\beta(x) = \tfrac{1}{1-x}$ , $\beta^2(x) = \tfrac{x-1}{x}$ , $\alpha(x) = \tfrac{1}{x}$ , $\alpha\beta(x) = 1-x$ and $\alpha\beta^2(x) = \tfrac{x}{x-1}$ form a group under composition, we can form (by substituting $x$ , $\beta(x)$ , $\beta^2(x)$ , $\alpha(x)$ , $\alpha\beta(x)$ and $\alpha\beta^2(x)$ for $x$ ) a $6\times6$ system of equations $\left(\begin{array}{cccccc} 1 & 0 & 0 & 2 & 0 & 3 \\ 0 & 1 & 0 & 3 & 2 & 0 \\ 0 &0 & 1 & 0 & 3 & 2 \\ 2 & 3 & 0 & 1 & 0 & 0 \\ 0 & 2 & 3 & 0 & 1 & 0 \\ 3 & 0 & 2 & 0 & 0 & 1 \end{array}\right) \left( \begin{array}{c} f(x) \\ f\big(\tfrac{1}{1-x}\big) \\ f\big(\tfrac{x-1}{x}\big) \\ f\big(\tfrac{1}{x}\big) \\ f(1-x) \\ f\big(\tfrac{x}{x-1}\big) \end{array} \right) \; = \; \left( \begin{array}{c} x \\ \tfrac{1}{1-x} \\ \tfrac{x-1}{x} \\ \tfrac{1}{x} \\ 1-x \\ \tfrac{x}{x-1} \end{array} \right)$ Inverting this matrix we obtain $\begin{aligned} f(x) & = \big( 1 \;\; 0 \;\; 0 \;\;0 \;\;0 \;\;0\big)\left(\begin{array}{cccccc} 1 & 0 & 0 & 2 & 0 & 3 \\ 0 & 1 & 0 & 3 & 2 & 0 \\ 0 &0 & 1 & 0 & 3 & 2 \\ 2 & 3 & 0 & 1 & 0 & 0 \\ 0 & 2 & 3 & 0 & 1 & 0 \\ 3 & 0 & 2 & 0 & 0 & 1 \end{array}\right)^{-1}\left( \begin{array}{c} x \\ \tfrac{1}{1-x} \\ \tfrac{x-1}{x} \\ \tfrac{1}{x} \\ 1-x \\ \tfrac{x}{x-1} \end{array} \right) \\ & = \frac{1}{24}\Big[-3x + \frac{1}{1-x} + \frac{x-1}{x} + \frac{3}{x} - 5(1-x) + 7\frac{x}{x-1}\Big] \; = \; \frac{2x^3 + x^2 + 5x - 2}{24x(x-1)} \end{aligned}$ and hence $f(4) = \tfrac{9}{16}$ , and so $16f(4) = \boxed{9}$ .