Linear Algebra Done Right · Problem 3D.24

The key insight is that $AB = I$ tells you $B$ is a right inverse of $A$, but you need to show it's also a left inverse. The trick is to use associativity of matrix multiplication strategically: multiply both sides of $AB = I$ on the left by $B$, then rearrange to isolate $BA$. This works because matrix multiplication is associative even though it's not commutative, so the grouping $(BA)B = B(AB)$ lets you substitute what you already know.

The key insight is that $AB = I$ tells you $B$ is a right inverse of $A$, but you need to show it's also a left inverse. The trick is to use associativity of matrix multiplication strategically: multiply both sides of $AB = I$ on the left by $B$, then rearrange to isolate $BA$. This works because matrix multiplication is associative even though it's not commutative, so the grouping $(BA)B = B(AB)$ lets you substitute what you already know.