Find $\left(f\circ g\right)\left(\left(x\text{,}y\right)\right)$.

Find $\left(g\circ f\right)\left(\left(x\text{,}y\right)\right)$.

State with a reason whether or not $f$ and $g$ commute.

Find the inverse of $f$.

$\left(f\circ g\right)\left(\left(x\text{,}y\right)\right)=f\left(g\left(\left(x\text{,}y\right)\right)\right)$  ($=f\left(\left(xy,x+y\right)\right)$)       (M1)

$=\left(xy+x+y,xy-x-y\right)$       A1A1

[3 marks]

$\left(g\circ f\right)\left(\left(x\text{,}y\right)\right)=g\left(f\left(\left(x\text{,}y\right)\right)\right)$

$=g\left(\left(x+y,x-y\right)\right)$

$=\left(\left(x+y\right)\left(x-y\right),x+y+x-y\right)$

$=\left(x^2-y^2\text{,}2x\right)$       A1A1

[2 marks]

no because $f\circ g\ne g\circ f$        R1

Note: Accept counter example.

[1 mark]

$f\left(\left(x\text{,}y\right)\right)=\left(a\text{,}b\right)\Rightarrow \left(x+y,x-y\right)=\left(a\text{,}b\right)$       (M1)

$\{\begin{array}{c}x=\frac{a+b}{2}\\ y=\frac{a-b}{2}\end{array}$       (M1)

$f^{-1}\left(\left(x\text{,}y\right)\right)=\left(\frac{x+y}{2},\frac{x-y}{2}\right)$        A1

[3 marks]