How to find a inverse of a multivariable function?

Question

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.

Answer 1

You have a linear function here, given by the matrix $$ \begin{bmatrix}3&-1\\1&-5 \end{bmatrix} $$ Can you invert the above?

Answer 2

You can split this into two separate functions $u, v:\Bbb R^2\to \Bbb R$ the following way: $$ u(x, y) = 3x-y\\ v(x, y) = x-5y $$ and we have $f(x, y) = (u(x, y), v(x, y))$. What we want is $x$ and $y$ expressed in terms of $u$ and $v$, i.e. solve the above set of equations for $x$ and $y$, so that you get two functions $x(u, v), y(u, v)$. Then $f^{-1}(u, v) = (x(u,v), y(u,v))$.