I have a function $f:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ defined as:
$$f(x,y) = (3x-y, x-5y)$$
I proved that it's a bijection, now I have to find the inverse function $f^{-1}$.
Because $f$ is a bijection, it has a inverse and this is true:
$$(f^{-1}\circ f)(x,y) = (x,y)$$
$$f^{-1}(3x-y,x-5) = (x,y)$$
I don't know where to go from here. In a one variable function I would do a substitution of the argument of $f^{-1}$ with a variable and express x with that variable, and then just switch places.
I tried to do a substitution like this:
$$3x-y = a$$
$$x-5y = b$$
And then express $x$ and $y$ by $a$ and $b$ , and get this:
$$f^{-1}(x,y) = (\frac{15x-3y}{42}, \frac{x-3y}{14})$$
But I'm not sure if I'm allowed to swap $x$ for $a$, and $y$ for $b$.
Any hint is highly appreciated.
Best Answer
You have a linear function here, given by the matrix $$ \begin{bmatrix}3&-1\\1&-5 \end{bmatrix} $$ Can you invert the above?