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Let A and B be two invertible matrices of order 3 x 3. If $det (ABA^{T})=8$ and $det (AB^{-1})=8$ , then $det (BA^{-1}B^{T})$ is equal to:
Option 1)
$\frac{1}{16}$
Option 2)
$\frac{1}{4}$
Option 3)
1
Option 4)
16

Answers (1)

best_answer

Property of inverse of a matrix -

Every invertible matrix possesses a unique inverse

-

Reversal law -

$\left ( AB \right )^{-1}=B^{-1}A^{-1}$

- wherein

$A$ and $B$ are invertible matrices of same order

Property of Transpose -

$\left ( \alpha A \right ){}'= \alpha A{}'$

$(AB)^{'} =B^{'}A^{'}$

- wherein

$\alpha$ being scalar ; $A{}'$ is transpose of A

$\left | A \right |^{2}\left | B \right |=8$ and $\frac{\left | A \right |}{\left | B \right |}=8$

$=>\left | A \right |=4$ and $=>{\left | B \right |}=\frac{1}{2}$

$\therefore \: det(BA^{-1}B^{T})=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

Option 1)
$\frac{1}{16}$
Option 2)

$\frac{1}{4}$

Option 3)

1

Option 4)

16

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