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

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

Explanation

Solution

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

Complete step by step solution:
From the definition of inverse of a matrix AA, we have AA−1=IA{A^{ - 1}} = I.
Since AA and BB are invertible matrices of the same order.
Therefore, apply the definition ofA−1{A^{ - 1}} to the matrix(AB)\left( {AB} \right);
(AB)(AB)−1=I\left( {AB} \right){\left( {AB} \right)^{ - 1}} = I
⇒A−1(AB)(AB)−1=A−1I\Rightarrow {A^{ - 1}}\left( {AB} \right){\left( {AB} \right)^{ - 1}} = {A^{ - 1}}I (Pre multiplying both sides by A−1{A^{ - 1}})
⇒(A−1A)B(AB)−1=A−1\Rightarrow \left( {{A^{ - 1}}A} \right)B{\left( {AB} \right)^{ - 1}} = {A^{ - 1}} (Since A−1I=A−1{A^{ - 1}}I = {A^{ - 1}})
⇒IB(AB)−1=A−1\Rightarrow IB{\left( {AB} \right)^{ - 1}} = {A^{ - 1}} (Since A−1A=I{A^{ - 1}}A = I)
⇒B(AB)−1=A−1\Rightarrow B{\left( {AB} \right)^{ - 1}} = {A^{ - 1}}
⇒B−1B(AB)−1=B−1A−1\Rightarrow {B^{ - 1}}B{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}} (Pre multiplying both sides by B−1{B^{ - 1}})
⇒I(AB)−1=B−1A−1\Rightarrow I{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}} (Since B−1B=I{B^{ - 1}}B = I)
⇒(AB)−1=B−1A−1\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×mm \times m.
(2) The inverse of a matrix is defined only for a non-singular matrix, whose determinant value does not equal to zero.