Problem. Let \(A,B\in M_{n\times n}(\mathbb{F})\) be such that \(AB=I_{n}\). Prove that \(A\) and \(B\) are inverses of each other.
Proof. As we know, the homogeneous system \(Bx=0\) has at least one solution (the trivial solution), hence we can mark the set of its solutions as \(U\), and definitionally \(\ker(T_{B})=U\).
Let \(x\in U\), and let us multiply both sides on the left by A:
\[A\cdot Bx=A\cdot0\therefore\quad A\cdot Bx=0\]
matrix multiplication is associative, hence \[AB\cdot x=0\therefore\quad I_{n}\cdot x=0\therefore\quad x=0\]
Which means \(\ker(T_{B})=\{0\}\).
Recall the theorem that \(T_{B}\) is injective if and only if its kernel is 0, which means \(T_{B}:\mathbb{\mathbb{F}}^{n}\rightarrow\mathbb{\mathbb{F}}^{n}\) is injective.
Furthermore, recall that a linear transformation between finite dimensional vector spaces of the same dimension is surjective if and only if it is injective. \(\mathbb{\mathbb{F}}^{n}\) is a finite dimensional vector space of dimension \(n\in\mathbb{N}\), therefore \(T_{B}\) is bijective, which means \(B\) is invertible, therefore B has an inverse \(B^{-1}\) and \(BB^{-1}=B^{-1}B=I_{n}\).
If we prove that \(B^{-1}=A\), we get \(BA=AB=I_{n}\), as required.
\[I_{n}B^{-1}=B^{-1}\underset{\text{substitution}}{\therefore}(AB)B^{-1}=B^{-1}\underset{\text{associativity}}{\therefore}A\left(BB^{-1}\right)=B^{-1}\]
\[\therefore AI_{n}=B^{-1}\therefore A=B^{-1}\] ∎
This is an exam problem I translated and solved. The original exam was written by the Linear Algebra 1 course lecturers at HUJI.