Question

Let \(B\) and \(C\) be two inverses of a square matrix \(A\). Which chain correctly proves that \(B=C\)?

Options

• \[B=BI=B(AC)=(BA)C=IA=A\]

• \[B=BI=B(AC)=(BA)C=IC=C\]

• \[B=BI=B(AC)=(AB)C=IC=C\]

• \[B=BI=B(CA)=(BA)C=IC=C\]

Answer 1

Since \(AC=I\), \(B(AC)=BI=B\). Also, \(BA=I\), so \((BA)C=IC=C\); therefore \(B=C\), proving uniqueness.