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Question: A is a \(3\times 3\) non-singular matrix with \[A{{A}^{1}}={{A}^{1}}A\text{ and B=}{{\text{A}}^{-1}}...

A is a 3×33\times 3 non-singular matrix with AA1=A1A and B=A−1A1A{{A}^{1}}={{A}^{1}}A\text{ and B=}{{\text{A}}^{-1}}{{A}^{1}} then BB1B{{B}^{1}} is equal to

& A.\text{ I+B} \\\ & \text{B}\text{. I} \\\ & \text{C}\text{. }{{\text{B}}^{-1}} \\\ & \text{D}\text{. }{{\left( {{B}^{-1}} \right)}^{1}} \\\ \end{aligned}$$
Explanation

Solution

We know that A−1{{A}^{-1}} is inverse of matrix and A1{{A}^{1}} is transpose of matrix. To solve this, first substitute the value of B=A−1A1\text{B=}{{\text{A}}^{-1}}{{A}^{1}} in BB1B{{B}^{1}} then use the property given in the question as AA1=A1AA{{A}^{1}}={{A}^{1}}A Try to take AA−1 and (A1)(A1)−1A{{A}^{-1}}\text{ and }\left( {{A}^{1}} \right){{\left( {{A}^{1}} \right)}^{-1}} together and use the fact that AA−1=IA{{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 A1{{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 3×33\times 3 matrix whose determinant is non zero.
Given that AA1=A1A . . . . . . . . . . . (i)A{{A}^{1}}={{A}^{1}}A\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}
Consider BB1B{{B}^{1}}
Given that B=A−1A1\text{B=}{{\text{A}}^{-1}}{{A}^{1}}
Substituting B=A−1A1 in BB1\text{B=}{{\text{A}}^{-1}}{{A}^{1}}\text{ in B}{{\text{B}}^{\text{1}}} we get:
BB1=(A−1A1)(A−1A1)1B{{B}^{1}}=\left( {{A}^{-1}}{{A}^{1}} \right){{\left( {{A}^{-1}}{{A}^{1}} \right)}^{1}}
Using the fact that (PQ)1=Q1P1{{\left( PQ \right)}^{1}}={{Q}^{1}}{{P}^{1}} in above we get:

& 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{aligned}$$ Using the fact that ${{\left( {{A}^{1}} \right)}^{1}}=A$ in above we get: $$\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{aligned} & 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{aligned}$$ So, the value of $$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.