Question

If $A$ and $B$ are invertible matrices of same order, then prove that ${\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$.

Answer 1

Hint:
If $A$ is a non-singular matrix, there is an existence of a matrix ${A^{ - 1}}$of same order, which is called the inverse matrix of$A$ such that it satisfies the property $A{A^{ - 1}} = {A^{ - 1}}A = I$, where $I$ is the identity matrix.

Complete step by step solution:
 From the definition of inverse of a matrix $A$, we have $A{A^{ - 1}} = I$.
Since $A$ and $B$ are invertible matrices of the same order.
Therefore, apply the definition of${A^{ - 1}}$ to the matrix$\left( {AB} \right)$;
$\left( {AB} \right){\left( {AB} \right)^{ - 1}} = I$
$ \Rightarrow {A^{ - 1}}\left( {AB} \right){\left( {AB} \right)^{ - 1}} = {A^{ - 1}}I$ (Pre multiplying both sides by ${A^{ - 1}}$)
$ \Rightarrow \left( {{A^{ - 1}}A} \right)B{\left( {AB} \right)^{ - 1}} = {A^{ - 1}}$ (Since ${A^{ - 1}}I = {A^{ - 1}}$)
$ \Rightarrow IB{\left( {AB} \right)^{ - 1}} = {A^{ - 1}}$ (Since ${A^{ - 1}}A = I$)
$ \Rightarrow B{\left( {AB} \right)^{ - 1}} = {A^{ - 1}}$
$ \Rightarrow {B^{ - 1}}B{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$ (Pre multiplying both sides by ${B^{ - 1}}$)
$ \Rightarrow I{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$ (Since ${B^{ - 1}}B = I$)
$ \Rightarrow {\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$

Note:
(1) The inverse of a matrix is defined only for a square matrix, whose order is $m \times m$.
(2) The inverse of a matrix is defined only for a non-singular matrix, whose determinant value does not equal to zero.