GATE Overflow Test Series | Linear Algebra | Test 1 | Question: 7

Question

If $A$ is an $n \times n$ non-singular matrix, such that $AA^{T} = A^{T}A$ and $B = A^{-1}A^{T},$ then $BB^{T}$ is equal to _____ (Mark all the appropriate choices)

• A. $I$
• B. $B^{-1}$
• C. $(B^{-1})^{T}$
• D. $I^{T}$

Answer 1

$AA^{T} = A^{T}A$ and $B = A^{-1}A^{T}$

$\implies BB^{T} = (A^{-1}A^{T})\;(A^{-1}A^{T})^{T}$

$\implies BB^{T} = (A^{-1}A^{T})\;(A^{T})^{T} (A^{-1})^{T}\;\quad \left[\because (AB)^{T} = B^{T} A^{T}\right]$

$\implies BB^{T} = (A^{-1}A^{T})\;A (A^{-1})^{T} \;\quad \left[\because (A^{T})^{T} = A\right]$

$ \implies BB^{T} = (A^{-1}A) A^{T}(A^{-1})^{T}  \quad \left [\because\;\text{Given that}:  AA^{T} = A^{T}A \right]$

$\implies BB^{T} = I A^{T}(A^{-1})^{T} = A^{T}(A^{-1})^{T} \quad \left [\because AA^{-1} = A^{-1}A = I \right]$

$\implies BB^{T} = (A^{-1}A)^{T} = I^{T} = I$

So, the correct answer is $A;D.$