Matrices and Determinants 3 Question 6
7. If $A$ is a $3 \times 3$ non-singular matrix such that $A A^{T}=A^{T} A$ and $B=A^{-1} A^{T}$, then $B B^{T}$ is equal to
(a) $I+B$
(b) $I$
(c) $B^{-1}$
(d) $\left(B^{-1}\right)^{T}$
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Answer:
Correct Answer: 7. (b)
Solution:
- PLAN: Use the following properties of transpose
$(A B)^{T}=B^{T} A^{T},\left(A^{T}\right)^{T}=A$ and $A^{-1} A=I$ and simplify. If $A$ is non-singular matrix, then $|A| \neq 0$.
Given, $\quad A A^{T}=A^{T} A$ and $B=A^{-1} A^{T}$
$ \begin{aligned} B B^{T} & =\left(A^{-1} A^{T}\right)\left(A^{-1} A^{T}\right)^{T} \\ & =A^{-1} A^{T} A\left(A^{-1}\right)^{T} \quad\left[\because(A B)^{T}=B^{T} A^{T}\right] \\ & =A^{-1} A A^{T}\left(A^{-1}\right)^{T} \quad\left[\because A A^{T}=A^{T} A\right] \\ & =I A^{T}\left(A^{-1}\right)^{T} \quad\left[\because A^{-1} A=I\right] \\ & =A^{T}\left(A^{-1}\right)^{T}=\left(A^{-1} A\right)^{T} \\ & =I^{T}=I \end{aligned} $
