Prove that the following function is onto and one-to-one $$f:\Bbb Z^2\to\Bbb Z^2:{n\brack m}\mapsto\begin{bmatrix}3&2\\4&3\end{bmatrix}{n\brack m}$$
it is onto because:
$$\left\{\begin{align*}&m = 9d - 4s\\
&n = 6d + 3s\end{align*}\right.$$
(after solving the system of equations)
onto because m and n are integers
one-to-one because:
$$\left\{\begin{align*}&3n + 2m = 3p + 2q\\
&4n + 3m = 4p + 3q\end{align*}\right.$$
which is equal to $L_2 - L_1$ and $L_1 - 2L_2$:
$$\left\{\begin{align*}&n = p\\
&n + m = p + q\end{align*}\right.$$
equal to:
$$\left\{\begin{align*}&n = p\\
&m = q\end{align*}\right.$$
I have to solve the system of equations for both (to prove it is onto and one-to-one) if I understood correctly?
 A: Yours can be interpreted as linear transformation/homomorphism, from $\Bbb Z^2$ to itself. 
Injectivity and surjectivity then boil down to the matrix being invertible or not, and the value of it's determinant. We have $$\begin{vmatrix}3&2\\4&3\end{vmatrix}=3\cdot 3 -4\cdot 2 =9-8=1$$
Since the determinant is $1$, the function is invertible (which means it is one-one and onto), and it's inverse matrix $$\left(\begin{matrix}3&-4\\-2&3\end{matrix}\right)$$ will also send integers to integers, which is indeed what we want.   
A: Let $f \colon X \to Y$, then:


*

*$f$ is onto (surjective) if and only if for every $y \in Y$, there exists some $x \in X$ such that $y = f(x)$.

*$f$ is one-to-one (injective) if and only if $$f(a) \neq f(b) \implies a \neq b$$ for every $a, b \in X$
Claim: $f$ is surjective.
Proof:
Observe that we have a matrix with $\textrm{Det}(M)=1$, then we know (from linear algebra) $M$ is invertible.
Let $$ M = \begin{pmatrix} 3 & 2 \\ 4 & 3 \end{pmatrix}.$$
Then $$ M^{-1} = \begin{pmatrix} 3 & -2 \\ -4 & 3 \end{pmatrix}.$$
Since $f(n,m) = (3n + 2m, 4n + 3m)$ and $f^{-1}(r,s) = (3r -2s, -4r + 3s)$, we know the map is surjective since for every $(r,s) \in \mathbb{Z}^{2}$, it is always true that $(r,s) = f(f^{-1}(r,s))$.
Lemma: Let $f \colon X \to Y$ such that $f^{-1}(f(x)) = x$ for every $x \in X$, then $f$ is injective.
Claim: $f$ is injective.
Proof:
Follows immediately from the above lemma, since it is easily verified that $f$ is left-invertible.
