Question

If $A$ and $B$ are two non-singular square matrices of order $n$, then prove that \[ (AB)^{-1} = B^{-1}A^{-1} \]

Hint:

Note: The order of multiplication reverses when taking the inverse of product of matrices.

Answer 1

Step 1: Recall definition.
For any non-singular matrix $M$, \[ M \cdot M^{-1} = I \]

Step 2: Take $M = AB$.
\[ (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} \] \[ = AIA^{-1} \] \[ = AA^{-1} = I \]

Step 3: Conclude.
Since $(AB)(B^{-1}A^{-1}) = I$, it follows that \[ (AB)^{-1} = B^{-1}A^{-1} \]

Final Answer: \[ \boxed{(AB)^{-1} = B^{-1}A^{-1}} \]