Question

A is a $3\times 3$ non-singular matrix with \[A{{A}^{1}}={{A}^{1}}A\text{ and B=}{{\text{A}}^{-1}}{{A}^{1}}\] then $B{{B}^{1}}$ is equal to
\[\begin{align}
  & A.\text{ I+B} \\
 & \text{B}\text{. I} \\
 & \text{C}\text{. }{{\text{B}}^{-1}} \\
 & \text{D}\text{. }{{\left( {{B}^{-1}} \right)}^{1}} \\
\end{align}\]

Answer 1

Hint: We know that ${{A}^{-1}}$ is inverse of matrix and ${{A}^{1}}$ is transpose of matrix. To solve this, first substitute the value of $\text{B=}{{\text{A}}^{-1}}{{A}^{1}}$ in $B{{B}^{1}}$ then use the property given in the question as $A{{A}^{1}}={{A}^{1}}A$ Try to take \[A{{A}^{-1}}\text{ and }\left( {{A}^{1}} \right){{\left( {{A}^{1}} \right)}^{-1}}\] together and use the fact that \[A{{A}^{-1}}=I\] to get result.

Complete step-by-step answer:
Transpose of a matrix is obtained by reversing rows and columns of a matrix ${{A}^{1}}$ is a transpose of a matrix A.
Non-singular matrix is a matrix whose determinant is non zero. So, we have A is a $3\times 3$ matrix whose determinant is non zero.
Given that \[A{{A}^{1}}={{A}^{1}}A\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Consider $B{{B}^{1}}$
Given that \[\text{B=}{{\text{A}}^{-1}}{{A}^{1}}\]
Substituting $\text{B=}{{\text{A}}^{-1}}{{A}^{1}}\text{ in B}{{\text{B}}^{\text{1}}}$ we get:
\[B{{B}^{1}}=\left( {{A}^{-1}}{{A}^{1}} \right){{\left( {{A}^{-1}}{{A}^{1}} \right)}^{1}}\]
Using the fact that ${{\left( PQ \right)}^{1}}={{Q}^{1}}{{P}^{1}}$ in above we get:
\[\begin{align}
  & B{{B}^{1}}=\left( {{A}^{-1}}{{A}^{1}} \right)\left( {{\left( {{A}^{1}} \right)}^{1}}{{\left( {{A}^{-1}} \right)}^{1}} \right) \\
 & \Rightarrow B{{B}^{1}}=\left( {{A}^{-1}}{{A}^{1}} \right)\left( {{\left( {{A}^{1}} \right)}^{1}}{{\left( {{A}^{-1}} \right)}^{1}} \right) \\
\end{align}\]
Using the fact that ${{\left( {{A}^{1}} \right)}^{1}}=A$ in above we get:
\[\Rightarrow B{{B}^{1}}=\left( {{A}^{-1}}{{A}^{1}} \right)\left( A{{\left( {{A}^{-1}} \right)}^{1}} \right)\]
Using the fact that ${{\left( {{A}^{-1}} \right)}^{1}}={{\left( {{A}^{1}} \right)}^{-1}}$ in above we get:
\[\Rightarrow B{{B}^{1}}=\left( {{A}^{-1}}{{A}^{1}} \right)\left( A{{\left( {{A}^{1}} \right)}^{-1}} \right)\]
Opening the bracket we get:
\[\Rightarrow B{{B}^{1}}=\left( {{A}^{-1}}{{A}^{1}}A{{\left( {{A}^{1}} \right)}^{-1}} \right)\]
Using equation (i) we have $A{{A}^{1}}={{A}^{1}}A$ in above we get:
\[\Rightarrow B{{B}^{1}}=\left( {{A}^{-1}}A{{A}^{1}}{{\left( {{A}^{1}} \right)}^{-1}} \right)\]
Now, using the fact that $A{{A}^{-1}}=I$
Where ${{A}^{-1}}$ is the inverse of a matrix, we get:
\[\begin{align}
  & B{{B}^{-1}}=\left( I\left( {{A}^{1}} \right){{\left( {{A}^{1}} \right)}^{-1}} \right) \\
 & \text{as }A{{A}^{-1}}=I\text{ then }\left( {{B}^{1}} \right){{\left( {{B}^{1}} \right)}^{-1}}=I \\
 & \Rightarrow B{{B}^{-1}}=\left( I-I \right) \\
 & \Rightarrow B{{B}^{-1}}=I \\
\end{align}\]
So, the value of \[B{{B}^{-1}}=I\]

So, the correct answer is “Option B”.

Note: The possibility of error in this question can be at the point where, we have to take the value of ${{A}^{1}}$ and inverse of ${{A}^{1}}$ here, \[{{\left( {{A}^{1}} \right)}^{-1}}{{A}^{1}}=I\text{ as }B{{B}^{-1}}=I\] where ${{B}^{-1}}$ is inverse of B.
Also, remember the matrix multiplication is not commutative, that is $AB\ne BA$ unless stated. So, we cannot use $A{{A}^{1}}={{A}^{1}}A$ unless stated in the question.