Let \(A\) and \(B\) be \(2 \times 2\) matrices such that \(AB = I\). Prove that \(BA = I\).
Solution
From \(AB=I_2\), taking determinants gives \[ \det(A)\det(B)=1. \] Hence \(\det(A)\ne0\), so \(A\) is invertible. Multiplying \(AB=I_2\) on the left by \(A^{-1}\) gives \[ B=A^{-1}. \] Therefore \[ BA=A^{-1}A=I_2. \]