First slide

Applications of determinant

Question

$Let A and B be two invertible matrices of order 3 \times 3 . If \det (A B A^{T}) = 8 and$ $\det (A B^{- 1}) = 8, then \det (B A^{- 1} B^{T}) is equal to$

Moderate

A

1
B

$\frac{1}{4}$
C

$\frac{1}{16}$
D

16

Solution

$Given, |A B A^{T}| = 8$
$\begin{aligned} \Rightarrow |A ∥ B ∥ A^{T}| = 8 [∵ | X Y | = | X | Y ∣] \\ ∴ | A |^{2} | B | = 8 \dots (i) [∵ |A^{T}| = | A |] \end{aligned}$
$\underline{\underline{Also}}, we have |A B^{- 1}| = 8 \Rightarrow |A ∥ B^{- 1}| = 8$
$\Rightarrow \frac{| A |}{| B |} = 8 ... (i i) [∵ | A^{- 1} | = | A |^{- 1} = \frac{1}{| A |}]$
On multiplying Eqs.(i) and (ii) , we get
$\begin{aligned} | A |^{3} = 8.8 = 4^{3} \\ \Rightarrow | A | = 4 \\ \Rightarrow | B | = \frac{| A |}{8} = \frac{4}{8} = \frac{1}{2} \end{aligned}$
$Now, |B A^{- 1} B^{T}| = | B | \frac{1}{| A |} | B | = (\frac{1}{2}) \frac{1}{4} (\frac{1}{2}) = \frac{1}{16}$

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